How do I perform the SQL Join equivalent in MongoDB?
How do I perform the SQL Join equivalent in MongoDB?
For example say you have two collections (users and comments) and I want to pull all the comments with pid=444 along with the user info for each.
comments
{ uid:12345, pid:444, comment="blah" }
{ uid:12345, pid:888, comment="asdf" }
{ uid:99999, pid:444, comment="qwer" }
users
{ uid:12345, name:"john" }
{ uid:99999, name:"mia" }
Is there a way to pull all the comments with a certain field (eg. ...find({pid:444}) ) and the user information associated with each comment in one go?
At the moment, I am first getting the comments which match my criteria, then figuring out all the uid's in that result set, getting the user objects, and merging them with the comment's results. Seems like I am doing it wrong.
join mongodb normalization
add a comment |
How do I perform the SQL Join equivalent in MongoDB?
For example say you have two collections (users and comments) and I want to pull all the comments with pid=444 along with the user info for each.
comments
{ uid:12345, pid:444, comment="blah" }
{ uid:12345, pid:888, comment="asdf" }
{ uid:99999, pid:444, comment="qwer" }
users
{ uid:12345, name:"john" }
{ uid:99999, name:"mia" }
Is there a way to pull all the comments with a certain field (eg. ...find({pid:444}) ) and the user information associated with each comment in one go?
At the moment, I am first getting the comments which match my criteria, then figuring out all the uid's in that result set, getting the user objects, and merging them with the comment's results. Seems like I am doing it wrong.
join mongodb normalization
30
The last answer on this question is probably the most relevant, since MongoDB 3.2+ implemented a join solution called $lookup. Thought I would push it here because maybe not everyone will read to the bottom. stackoverflow.com/a/33511166/2593330
– thefourtheye
Nov 22 '15 at 16:33
6
Correct, $lookup was introduced in MongoDB 3.2. Details can be found at docs.mongodb.org/master/reference/operator/aggregation/lookup/…
– NDB
Nov 23 '15 at 13:20
i wonder why do not accept this anwser or any comment: stackoverflow.com/a/33511166/6941294
– Nozar Safari
Dec 18 '17 at 6:26
add a comment |
How do I perform the SQL Join equivalent in MongoDB?
For example say you have two collections (users and comments) and I want to pull all the comments with pid=444 along with the user info for each.
comments
{ uid:12345, pid:444, comment="blah" }
{ uid:12345, pid:888, comment="asdf" }
{ uid:99999, pid:444, comment="qwer" }
users
{ uid:12345, name:"john" }
{ uid:99999, name:"mia" }
Is there a way to pull all the comments with a certain field (eg. ...find({pid:444}) ) and the user information associated with each comment in one go?
At the moment, I am first getting the comments which match my criteria, then figuring out all the uid's in that result set, getting the user objects, and merging them with the comment's results. Seems like I am doing it wrong.
join mongodb normalization
How do I perform the SQL Join equivalent in MongoDB?
For example say you have two collections (users and comments) and I want to pull all the comments with pid=444 along with the user info for each.
comments
{ uid:12345, pid:444, comment="blah" }
{ uid:12345, pid:888, comment="asdf" }
{ uid:99999, pid:444, comment="qwer" }
users
{ uid:12345, name:"john" }
{ uid:99999, name:"mia" }
Is there a way to pull all the comments with a certain field (eg. ...find({pid:444}) ) and the user information associated with each comment in one go?
At the moment, I am first getting the comments which match my criteria, then figuring out all the uid's in that result set, getting the user objects, and merging them with the comment's results. Seems like I am doing it wrong.
join mongodb normalization
join mongodb normalization
edited Jun 19 '12 at 11:55
Ozair Kafray
12.1k54672
12.1k54672
asked Feb 28 '10 at 8:15
The UnknownThe Unknown
8,364266489
8,364266489
30
The last answer on this question is probably the most relevant, since MongoDB 3.2+ implemented a join solution called $lookup. Thought I would push it here because maybe not everyone will read to the bottom. stackoverflow.com/a/33511166/2593330
– thefourtheye
Nov 22 '15 at 16:33
6
Correct, $lookup was introduced in MongoDB 3.2. Details can be found at docs.mongodb.org/master/reference/operator/aggregation/lookup/…
– NDB
Nov 23 '15 at 13:20
i wonder why do not accept this anwser or any comment: stackoverflow.com/a/33511166/6941294
– Nozar Safari
Dec 18 '17 at 6:26
add a comment |
30
The last answer on this question is probably the most relevant, since MongoDB 3.2+ implemented a join solution called $lookup. Thought I would push it here because maybe not everyone will read to the bottom. stackoverflow.com/a/33511166/2593330
– thefourtheye
Nov 22 '15 at 16:33
6
Correct, $lookup was introduced in MongoDB 3.2. Details can be found at docs.mongodb.org/master/reference/operator/aggregation/lookup/…
– NDB
Nov 23 '15 at 13:20
i wonder why do not accept this anwser or any comment: stackoverflow.com/a/33511166/6941294
– Nozar Safari
Dec 18 '17 at 6:26
30
30
The last answer on this question is probably the most relevant, since MongoDB 3.2+ implemented a join solution called $lookup. Thought I would push it here because maybe not everyone will read to the bottom. stackoverflow.com/a/33511166/2593330
– thefourtheye
Nov 22 '15 at 16:33
The last answer on this question is probably the most relevant, since MongoDB 3.2+ implemented a join solution called $lookup. Thought I would push it here because maybe not everyone will read to the bottom. stackoverflow.com/a/33511166/2593330
– thefourtheye
Nov 22 '15 at 16:33
6
6
Correct, $lookup was introduced in MongoDB 3.2. Details can be found at docs.mongodb.org/master/reference/operator/aggregation/lookup/…
– NDB
Nov 23 '15 at 13:20
Correct, $lookup was introduced in MongoDB 3.2. Details can be found at docs.mongodb.org/master/reference/operator/aggregation/lookup/…
– NDB
Nov 23 '15 at 13:20
i wonder why do not accept this anwser or any comment: stackoverflow.com/a/33511166/6941294
– Nozar Safari
Dec 18 '17 at 6:26
i wonder why do not accept this anwser or any comment: stackoverflow.com/a/33511166/6941294
– Nozar Safari
Dec 18 '17 at 6:26
add a comment |
19 Answers
19
active
oldest
votes
As of Mongo 3.2 the answers to this question are mostly no longer correct. The new $lookup operator added to the aggregation pipeline is essentially identical to a left outer join:
https://docs.mongodb.org/master/reference/operator/aggregation/lookup/#pipe._S_lookup
From the docs:
{
$lookup:
{
from: <collection to join>,
localField: <field from the input documents>,
foreignField: <field from the documents of the "from" collection>,
as: <output array field>
}
}
Of course Mongo is not a relational database, and the devs are being careful to recommend specific use cases for $lookup, but at least as of 3.2 doing join is now possible with MongoDB.
11
is $lookup performance-wise the better option?
– Mohsen Shakiba
Nov 21 '15 at 9:06
5
Agreed. This is the best answer.
– NDB
Nov 23 '15 at 13:21
@clayton : How about more then two collections?
– Dipen Dedania
Jan 8 '16 at 10:36
1
@DipenDedania just add additional $lookup stages to the aggregation pipeline.
– Clayton Gulick
Jan 9 '16 at 1:01
I cant join any field in array in left collection with its corresponding id in right collection.can anybody help me??
– Prateek Singh
Feb 1 '16 at 16:52
|
show 5 more comments
This page on the official mongodb site addresses exactly this question:
http://docs.mongodb.org/ecosystem/tutorial/model-data-for-ruby-on-rails/
When we display our list of stories, we'll need to show the name of the user who posted the story. If we were using a relational database, we could perform a join on users and stores, and get all our objects in a single query. But MongoDB does not support joins and so, at times, requires bit of denormalization. Here, this means caching the 'username' attribute.
Relational purists may be feeling uneasy already, as if we were violating some universal law. But let’s bear in mind that MongoDB collections are not equivalent to relational tables; each serves a unique design objective. A normalized table provides an atomic, isolated chunk of data. A document, however, more closely represents an object as a whole. In the case of a social news site, it can be argued that a username is intrinsic to the story being posted.
114
Then why so many people love MongoDB. I dont get it :|
– Boris D. Teoharov
Sep 15 '13 at 15:47
48
@dudelgrincen it's a paradigm shift from normalization and relational databases. The goal of a NoSQL is to read and write from the database very quickly. With BigData you're going to have scads of application and front end servers with lower numbers on DBs. You're expected to do millions of transactions a second. Offload the heavy lifting from the database and put it onto the application level. If you need deep analysis, you run an integration job that puts your data into an OLAP database. You shouldn't be getting many deep queries from your OLTP dbs anyway.
– Snowburnt
Nov 4 '13 at 1:53
17
@dudelgrincen I should also say that it's not for every project or design. If you have something that works in a SQL type database why change it? If you can't massage your schema to work with noSQL, then don't.
– Snowburnt
Nov 12 '13 at 0:30
8
Migrations and a constantly evolving schemas are also a lot easier to manage on a NoSQL system.
– justin
May 6 '14 at 20:09
11
What if the user has 3.540 posts in the website, and he does change his username in profile? Should every post be updated with the new username?
– Ivo Pereira
Mar 2 '16 at 17:39
|
show 7 more comments
We can merge/join all data inside only one collection with a easy function in few lines using the mongodb client console, and now we could be able of perform the desired query.
Below a complete example,
.- Authors:
db.authors.insert([
{
_id: 'a1',
name: { first: 'orlando', last: 'becerra' },
age: 27
},
{
_id: 'a2',
name: { first: 'mayra', last: 'sanchez' },
age: 21
}
]);
.- Categories:
db.categories.insert([
{
_id: 'c1',
name: 'sci-fi'
},
{
_id: 'c2',
name: 'romance'
}
]);
.- Books
db.books.insert([
{
_id: 'b1',
name: 'Groovy Book',
category: 'c1',
authors: ['a1']
},
{
_id: 'b2',
name: 'Java Book',
category: 'c2',
authors: ['a1','a2']
},
]);
.- Book lending
db.lendings.insert([
{
_id: 'l1',
book: 'b1',
date: new Date('01/01/11'),
lendingBy: 'jose'
},
{
_id: 'l2',
book: 'b1',
date: new Date('02/02/12'),
lendingBy: 'maria'
}
]);
.- The magic:
db.books.find().forEach(
function (newBook) {
newBook.category = db.categories.findOne( { "_id": newBook.category } );
newBook.lendings = db.lendings.find( { "book": newBook._id } ).toArray();
newBook.authors = db.authors.find( { "_id": { $in: newBook.authors } } ).toArray();
db.booksReloaded.insert(newBook);
}
);
.- Get the new collection data:
db.booksReloaded.find().pretty()
.- Response :)
{
"_id" : "b1",
"name" : "Groovy Book",
"category" : {
"_id" : "c1",
"name" : "sci-fi"
},
"authors" : [
{
"_id" : "a1",
"name" : {
"first" : "orlando",
"last" : "becerra"
},
"age" : 27
}
],
"lendings" : [
{
"_id" : "l1",
"book" : "b1",
"date" : ISODate("2011-01-01T00:00:00Z"),
"lendingBy" : "jose"
},
{
"_id" : "l2",
"book" : "b1",
"date" : ISODate("2012-02-02T00:00:00Z"),
"lendingBy" : "maria"
}
]
}
{
"_id" : "b2",
"name" : "Java Book",
"category" : {
"_id" : "c2",
"name" : "romance"
},
"authors" : [
{
"_id" : "a1",
"name" : {
"first" : "orlando",
"last" : "becerra"
},
"age" : 27
},
{
"_id" : "a2",
"name" : {
"first" : "mayra",
"last" : "sanchez"
},
"age" : 21
}
],
"lendings" : [ ]
}
I hope this lines can help you.
2
i'm wondering if this same code can be ran using doctrine mongodb?
– abbood
May 30 '14 at 13:46
4
What happens when one of the references objects gets an update? Does that update automatically reflect in the book object? Or does that loop need to run again?
– balupton
Jun 4 '14 at 5:55
9
This is fine as long as your data is small. It is going to bring each book content to your client and then fetch each category, lending and authors one by one. The moment your books are in thousands, this would go really really slow. A better technique probably would be to use aggregation pipeline and output the merged data into a separate collection. Let me get back to it again. I will add that an answer.
– Sandeep Giri
Jun 19 '14 at 15:31
Can you adapt your algorithm to this other example? stackoverflow.com/q/32718079/287948
– Peter Krauss
Sep 22 '15 at 14:13
1
@SandeepGiri how can i do the aggregate pipeline since i have really really intensive data in separated collection need join ??
– Yassine Abdul-Rahman
Oct 7 '15 at 20:16
|
show 3 more comments
You have to do it the way you described. MongoDB is a non-relational database and doesn't support joins.
4
Seems wrong performance wise coming from a sql server background, but its maybe not that bad with a document db?
– terjetyl
Jul 15 '10 at 18:20
3
from a sql server background as well, I would appreciate MongoDB taking a 'result set' (with selected returned fields) as input for a new query in one go, much like nested queries in SQL
– Stijn Sanders
Nov 26 '10 at 23:17
1
@terjetyl You have to really plan for it. What fields are you going be presenting on the front end, if it's a limited amount in an individual view then you take those as embedded documents. The key is to not need to do joins. If you want to do deep analysis, you do it after the fact in another database. Run a job that transforms the data into an OLAP cube for optimal performance.
– Snowburnt
Nov 4 '13 at 1:56
3
From mongo 3.2 version left joins are supported.
– Somnath Muluk
Nov 26 '15 at 11:12
add a comment |
Here's an example of a "join" * Actors and Movies collections:
https://github.com/mongodb/cookbook/blob/master/content/patterns/pivot.txt
It makes use of .mapReduce()
method
* join - an alternative to join in document-oriented databases
16
-1, This is NOT joining data from two collections. It is using data from a single collection (actors) pivoting data around. So that things that were keys are now values and values are now keys... very different than a JOIN.
– Evan Teran
May 22 '12 at 17:44
12
This exactly what you have to do, MongoDB is not relational but document oriented. MapReduce allows to play with data with big performance (you can use cluster etc....) but even for simple cases, its very useful !
– Thomas Decaux
Jun 17 '12 at 19:16
Yet the best approach. Very useful thanks
– Singh
Oct 15 '15 at 13:45
add a comment |
As others have pointed out you are trying to create a relational database from none relational database which you really don't want to do but anyways, if you have a case that you have to do this here is a solution you can use. We first do a foreach find on collection A( or in your case users) and then we get each item as an object then we use object property (in your case uid) to lookup in our second collection (in your case comments) if we can find it then we have a match and we can print or do something with it.
Hope this helps you and good luck :)
db.users.find().forEach(
function (object) {
var commonInBoth=db.comments.findOne({ "uid": object.uid} );
if (commonInBoth != null) {
printjson(commonInBoth) ;
printjson(object) ;
}else {
// did not match so we don't care in this case
}
});
1
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Wont this find the item we are currently looping on?
– Skarlinski
Mar 29 '17 at 12:58
This is poor performance wise for large collections.
– Kusaasira Joshua
Feb 7 at 10:08
add a comment |
It depends on what you're trying to do.
You currently have it set up as a normalized database, which is fine, and the way you are doing it is appropriate.
However, there are other ways of doing it.
You could have a posts collection that has imbedded comments for each post with references to the users that you can iteratively query to get. You could store the user's name with the comments, you could store them all in one document.
The thing with NoSQL is it's designed for flexible schemas and very fast reading and writing. In a typical Big Data farm the database is the biggest bottleneck, you have fewer database engines than you do application and front end servers...they're more expensive but more powerful, also hard drive space is very cheap comparatively. Normalization comes from the concept of trying to save space, but it comes with a cost at making your databases perform complicated Joins and verifying the integrity of relationships, performing cascading operations. All of which saves the developers some headaches if they designed the database properly.
With NoSQL, if you accept that redundancy and storage space aren't issues because of their cost (both in processor time required to do updates and hard drive costs to store extra data), denormalizing isn't an issue (for embedded arrays that become hundreds of thousands of items it can be a performance issue, but most of the time that's not a problem). Additionally you'll have several application and front end servers for every database cluster. Have them do the heavy lifting of the joins and let the database servers stick to reading and writing.
TL;DR: What you're doing is fine, and there are other ways of doing it. Check out the mongodb documentation's data model patterns for some great examples. http://docs.mongodb.org/manual/data-modeling/
7
"Normalization comes from the concept of trying to save space" I question this. IMHO normalization comes from the concept of avoiding redundancy. Say you store the name of a user along with a blogpost. What if she marries? In a not normalized model you will have to wade through all posts and change the name. In a normalized model you usually change ONE record.
– DanielKhan
Nov 27 '13 at 13:28
@DanielKhan preventing redundancy and saving space are similar concepts, but on re-analysis I do agree, redundancy is the root cause for this design. I'll reword. Thanks for the note.
– Snowburnt
Nov 27 '13 at 19:15
add a comment |
There is a specification that a lot of drivers support that's called DBRef.
DBRef is a more formal specification for creating references between documents. DBRefs (generally) include a collection name as well as an object id. Most developers only use DBRefs if the collection can change from one document to the next. If your referenced collection will always be the same, the manual references outlined above are more efficient.
Taken from MongoDB Documentation: Data Models > Data Model Reference >
Database References
add a comment |
You can join two collection in Mongo by using lookup which is offered in 3.2 version. In your case the query would be
db.comments.aggregate({
$lookup:{
from:"users",
localField:"uid",
foreignField:"uid",
as:"users_comments"
}
})
or you can also join with respect to users then there will be a little change as given below.
db.users.aggregate({
$lookup:{
from:"comments",
localField:"uid",
foreignField:"uid",
as:"users_comments"
}
})
It will work just same as left and right join in SQL.
add a comment |
With right combination of $lookup, $project and $match, you can join mutiple tables on multiple parameters. This is because they can be chained multiple times.
Suppose we want to do following (reference)
SELECT S.* FROM LeftTable S
LEFT JOIN RightTable R ON S.ID =R.ID AND S.MID =R.MID WHERE R.TIM >0 AND
S.MOB IS NOT NULL
Step 1: Link all tables
you can $lookup as many tables as you want.
$lookup - one for each table in query
$unwind - because data is denormalised correctly, else wrapped in arrays
Python code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "R"}
])
Step 2: Define all conditionals
$project : define all conditional statements here, plus all the variables you'd like to select.
Python Code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "R"},
# define conditionals + variables
{"$project": {
"midEq": {"$eq": ["$MID", "$R.MID"]},
"ID": 1, "MOB": 1, "MID": 1
}}
])
Step 3: Join all the conditionals
$match - join all conditions using OR or AND etc. There can be multiples of these.
$project: undefine all conditionals
Python Code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "$R"},
# define conditionals + variables
{"$project": {
"midEq": {"$eq": ["$MID", "$R.MID"]},
"ID": 1, "MOB": 1, "MID": 1
}},
# join all conditionals
{"$match": {
"$and": [
{"R.TIM": {"$gt": 0}},
{"MOB": {"$exists": True}},
{"midEq": {"$eq": True}}
]}},
# undefine conditionals
{"$project": {
"midEq": 0
}}
])
Pretty much any combination of tables, conditionals and joins can be done in this manner.
add a comment |
Before 3.2.6, Mongodb does not support join query as like mysql. below solution which works for you.
db.getCollection('comments').aggregate([
{$match : {pid : 444}},
{$lookup: {from: "users",localField: "uid",foreignField: "uid",as: "userData"}},
])
add a comment |
You can run SQL queries including join on MongoDB with mongo_fdw from Postgres.
add a comment |
$lookup (aggregation)
Performs a left outer join to an unsharded collection in the same database to filter in documents from the “joined” collection for processing. To each input document, the $lookup stage adds a new array field whose elements are the matching documents from the “joined” collection. The $lookup stage passes these reshaped documents to the next stage.
The $lookup stage has the following syntaxes:
Equality Match
To perform an equality match between a field from the input documents with a field from the documents of the “joined” collection, the $lookup stage has the following syntax:
{
$lookup:
{
from: <collection to join>,
localField: <field from the input documents>,
foreignField: <field from the documents of the "from" collection>,
as: <output array field>
}
}
The operation would correspond to the following pseudo-SQL statement:
SELECT *, <output array field>
FROM collection
WHERE <output array field> IN (SELECT <documents as determined from the pipeline>
FROM <collection to join>
WHERE <pipeline> );
Mongo URL
add a comment |
MongoDB does not allow joins, but you can use plugins to handle that. Check the mongo-join plugin. It's the best and I have already used it. You can install it using npm directly like this npm install mongo-join
. You can check out the full documentation with examples.
(++) really helpful tool when we need to join (N) collections
(--) we can apply conditions just on the top level of the query
Example
var Join = require('mongo-join').Join, mongodb = require('mongodb'), Db = mongodb.Db, Server = mongodb.Server;
db.open(function (err, Database) {
Database.collection('Appoint', function (err, Appoints) {
/* we can put conditions just on the top level */
Appoints.find({_id_Doctor: id_doctor ,full_date :{ $gte: start_date },
full_date :{ $lte: end_date }}, function (err, cursor) {
var join = new Join(Database).on({
field: '_id_Doctor', // <- field in Appoints document
to: '_id', // <- field in User doc. treated as ObjectID automatically.
from: 'User' // <- collection name for User doc
}).on({
field: '_id_Patient', // <- field in Appoints doc
to: '_id', // <- field in User doc. treated as ObjectID automatically.
from: 'User' // <- collection name for User doc
})
join.toArray(cursor, function (err, joinedDocs) {
/* do what ever you want here */
/* you can fetch the table and apply your own conditions */
.....
.....
.....
resp.status(200);
resp.json({
"status": 200,
"message": "success",
"Appoints_Range": joinedDocs,
});
return resp;
});
});
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Indeed I upvoted due to is a solution
– Maga
Oct 29 '15 at 10:21
add a comment |
playORM can do it for you using S-SQL(Scalable SQL) which just adds partitioning such that you can do joins within partitions.
add a comment |
You can do it using the aggregation pipeline, but it's a pain to write it yourself.
You can use mongo-join-query
to create the aggregation pipeline automatically from your query.
This is how your query would look like:
const mongoose = require("mongoose");
const joinQuery = require("mongo-join-query");
joinQuery(
mongoose.models.Comment,
{
find: { pid:444 },
populate: ["uid"]
},
(err, res) => (err ? console.log("Error:", err) : console.log("Success:", res.results))
);
Your result would have the user object in the uid
field and you can link as many levels deep as you want. You can populate the reference to the user, which makes reference to a Team, which makes reference to something else, etc..
Disclaimer: I wrote mongo-join-query
to tackle this exact problem.
add a comment |
Nope, it doesn't seem like you're doing it wrong. MongoDB joins are "client side". Pretty much like you said:
At the moment, I am first getting the comments which match my criteria, then figuring out all the uid's in that result set, getting the user objects, and merging them with the comment's results. Seems like I am doing it wrong.
1) Select from the collection you're interested in.
2) From that collection pull out ID's you need
3) Select from other collections
4) Decorate your original results.
It's not a "real" join, but it's actually alot more useful than a SQL join because you don't have to deal with duplicate rows for "many" sided joins, instead your decorating the originally selected set.
There is alot of nonsense and FUD on this page. Turns out 5 years later MongoDB is still a thing.
'you don't have to deal with duplicate rows for "many" sided joins' - no idea what you mean by this. Can you clarify?
– Mark Amery
Sep 20 '15 at 20:23
Downvote, really?
– Michael Cole
Sep 21 '15 at 14:46
1
@MarkAmery, sure. In SQL a n-n relationship will return duplicate rows. E.g. Friends. If Bob is friends with Mary and Jane, you'll get 2 rows for Bob: Bob,Mary and Bob,Jane. 2 Bobs is a lie, there is only one Bob. With client-side joins you can start with Bob and decorate how you like: Bob, "Mary and Jane". SQL let's you do this with subqueries, but that's doing work on the db server that could be done on the client.
– Michael Cole
Sep 21 '15 at 14:51
add a comment |
I think, if You need normalized data tables - You need to try some other database solutions.
But I've foun that sollution for MOngo on Git
By the way, in inserts code - it has movie's name, but noi movie's ID.
Problem
You have a collection of Actors with an array of the Movies they've done.
You want to generate a collection of Movies with an array of Actors in each.
Some sample data
db.actors.insert( { actor: "Richard Gere", movies: ['Pretty Woman', 'Runaway Bride', 'Chicago'] });
db.actors.insert( { actor: "Julia Roberts", movies: ['Pretty Woman', 'Runaway Bride', 'Erin Brockovich'] });
Solution
We need to loop through each movie in the Actor document and emit each Movie individually.
The catch here is in the reduce phase. We cannot emit an array from the reduce phase, so we must build an Actors array inside of the "value" document that is returned.
The code
map = function() {
for(var i in this.movies){
key = { movie: this.movies[i] };
value = { actors: [ this.actor ] };
emit(key, value);
}
}
reduce = function(key, values) {
actor_list = { actors: };
for(var i in values) {
actor_list.actors = values[i].actors.concat(actor_list.actors);
}
return actor_list;
}
Notice how actor_list is actually a javascript object that contains an array. Also notice that map emits the same structure.
Run the following to execute the map / reduce, output it to the "pivot" collection and print the result:
printjson(db.actors.mapReduce(map, reduce, "pivot"));
db.pivot.find().forEach(printjson);
Here is the sample output, note that "Pretty Woman" and "Runaway Bride" have both "Richard Gere" and "Julia Roberts".
{ "_id" : { "movie" : "Chicago" }, "value" : { "actors" : [ "Richard Gere" ] } }
{ "_id" : { "movie" : "Erin Brockovich" }, "value" : { "actors" : [ "Julia Roberts" ] } }
{ "_id" : { "movie" : "Pretty Woman" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
{ "_id" : { "movie" : "Runaway Bride" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
Note that most of the content of this answer (i.e. the bit that's in comprehensible English) is copied from the MongoDB cookbook at the GitHub link the answerer provided.
– Mark Amery
Oct 5 '15 at 14:23
add a comment |
We can merge two collection by using mongoDB sub query. Here is example,
Commentss--
`db.commentss.insert([
{ uid:12345, pid:444, comment:"blah" },
{ uid:12345, pid:888, comment:"asdf" },
{ uid:99999, pid:444, comment:"qwer" }])`
Userss--
db.userss.insert([
{ uid:12345, name:"john" },
{ uid:99999, name:"mia" }])
MongoDB sub query for JOIN--
`db.commentss.find().forEach(
function (newComments) {
newComments.userss = db.userss.find( { "uid": newComments.uid } ).toArray();
db.newCommentUsers.insert(newComments);
}
);`
Get result from newly generated Collection--
db.newCommentUsers.find().pretty()
Result--
`{
"_id" : ObjectId("5511236e29709afa03f226ef"),
"uid" : 12345,
"pid" : 444,
"comment" : "blah",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f2"),
"uid" : 12345,
"name" : "john"
}
]
}
{
"_id" : ObjectId("5511236e29709afa03f226f0"),
"uid" : 12345,
"pid" : 888,
"comment" : "asdf",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f2"),
"uid" : 12345,
"name" : "john"
}
]
}
{
"_id" : ObjectId("5511236e29709afa03f226f1"),
"uid" : 99999,
"pid" : 444,
"comment" : "qwer",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f3"),
"uid" : 99999,
"name" : "mia"
}
]
}`
Hope so this will help.
7
Why did you basically copy this nearly identical, one-year-old answer? stackoverflow.com/a/22739813/4186945
– hackel
May 5 '15 at 20:45
add a comment |
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As of Mongo 3.2 the answers to this question are mostly no longer correct. The new $lookup operator added to the aggregation pipeline is essentially identical to a left outer join:
https://docs.mongodb.org/master/reference/operator/aggregation/lookup/#pipe._S_lookup
From the docs:
{
$lookup:
{
from: <collection to join>,
localField: <field from the input documents>,
foreignField: <field from the documents of the "from" collection>,
as: <output array field>
}
}
Of course Mongo is not a relational database, and the devs are being careful to recommend specific use cases for $lookup, but at least as of 3.2 doing join is now possible with MongoDB.
11
is $lookup performance-wise the better option?
– Mohsen Shakiba
Nov 21 '15 at 9:06
5
Agreed. This is the best answer.
– NDB
Nov 23 '15 at 13:21
@clayton : How about more then two collections?
– Dipen Dedania
Jan 8 '16 at 10:36
1
@DipenDedania just add additional $lookup stages to the aggregation pipeline.
– Clayton Gulick
Jan 9 '16 at 1:01
I cant join any field in array in left collection with its corresponding id in right collection.can anybody help me??
– Prateek Singh
Feb 1 '16 at 16:52
|
show 5 more comments
As of Mongo 3.2 the answers to this question are mostly no longer correct. The new $lookup operator added to the aggregation pipeline is essentially identical to a left outer join:
https://docs.mongodb.org/master/reference/operator/aggregation/lookup/#pipe._S_lookup
From the docs:
{
$lookup:
{
from: <collection to join>,
localField: <field from the input documents>,
foreignField: <field from the documents of the "from" collection>,
as: <output array field>
}
}
Of course Mongo is not a relational database, and the devs are being careful to recommend specific use cases for $lookup, but at least as of 3.2 doing join is now possible with MongoDB.
11
is $lookup performance-wise the better option?
– Mohsen Shakiba
Nov 21 '15 at 9:06
5
Agreed. This is the best answer.
– NDB
Nov 23 '15 at 13:21
@clayton : How about more then two collections?
– Dipen Dedania
Jan 8 '16 at 10:36
1
@DipenDedania just add additional $lookup stages to the aggregation pipeline.
– Clayton Gulick
Jan 9 '16 at 1:01
I cant join any field in array in left collection with its corresponding id in right collection.can anybody help me??
– Prateek Singh
Feb 1 '16 at 16:52
|
show 5 more comments
As of Mongo 3.2 the answers to this question are mostly no longer correct. The new $lookup operator added to the aggregation pipeline is essentially identical to a left outer join:
https://docs.mongodb.org/master/reference/operator/aggregation/lookup/#pipe._S_lookup
From the docs:
{
$lookup:
{
from: <collection to join>,
localField: <field from the input documents>,
foreignField: <field from the documents of the "from" collection>,
as: <output array field>
}
}
Of course Mongo is not a relational database, and the devs are being careful to recommend specific use cases for $lookup, but at least as of 3.2 doing join is now possible with MongoDB.
As of Mongo 3.2 the answers to this question are mostly no longer correct. The new $lookup operator added to the aggregation pipeline is essentially identical to a left outer join:
https://docs.mongodb.org/master/reference/operator/aggregation/lookup/#pipe._S_lookup
From the docs:
{
$lookup:
{
from: <collection to join>,
localField: <field from the input documents>,
foreignField: <field from the documents of the "from" collection>,
as: <output array field>
}
}
Of course Mongo is not a relational database, and the devs are being careful to recommend specific use cases for $lookup, but at least as of 3.2 doing join is now possible with MongoDB.
edited Apr 10 '17 at 5:25
Ahmad Baktash Hayeri
4,61722237
4,61722237
answered Nov 3 '15 at 23:35
Clayton GulickClayton Gulick
6,61722621
6,61722621
11
is $lookup performance-wise the better option?
– Mohsen Shakiba
Nov 21 '15 at 9:06
5
Agreed. This is the best answer.
– NDB
Nov 23 '15 at 13:21
@clayton : How about more then two collections?
– Dipen Dedania
Jan 8 '16 at 10:36
1
@DipenDedania just add additional $lookup stages to the aggregation pipeline.
– Clayton Gulick
Jan 9 '16 at 1:01
I cant join any field in array in left collection with its corresponding id in right collection.can anybody help me??
– Prateek Singh
Feb 1 '16 at 16:52
|
show 5 more comments
11
is $lookup performance-wise the better option?
– Mohsen Shakiba
Nov 21 '15 at 9:06
5
Agreed. This is the best answer.
– NDB
Nov 23 '15 at 13:21
@clayton : How about more then two collections?
– Dipen Dedania
Jan 8 '16 at 10:36
1
@DipenDedania just add additional $lookup stages to the aggregation pipeline.
– Clayton Gulick
Jan 9 '16 at 1:01
I cant join any field in array in left collection with its corresponding id in right collection.can anybody help me??
– Prateek Singh
Feb 1 '16 at 16:52
11
11
is $lookup performance-wise the better option?
– Mohsen Shakiba
Nov 21 '15 at 9:06
is $lookup performance-wise the better option?
– Mohsen Shakiba
Nov 21 '15 at 9:06
5
5
Agreed. This is the best answer.
– NDB
Nov 23 '15 at 13:21
Agreed. This is the best answer.
– NDB
Nov 23 '15 at 13:21
@clayton : How about more then two collections?
– Dipen Dedania
Jan 8 '16 at 10:36
@clayton : How about more then two collections?
– Dipen Dedania
Jan 8 '16 at 10:36
1
1
@DipenDedania just add additional $lookup stages to the aggregation pipeline.
– Clayton Gulick
Jan 9 '16 at 1:01
@DipenDedania just add additional $lookup stages to the aggregation pipeline.
– Clayton Gulick
Jan 9 '16 at 1:01
I cant join any field in array in left collection with its corresponding id in right collection.can anybody help me??
– Prateek Singh
Feb 1 '16 at 16:52
I cant join any field in array in left collection with its corresponding id in right collection.can anybody help me??
– Prateek Singh
Feb 1 '16 at 16:52
|
show 5 more comments
This page on the official mongodb site addresses exactly this question:
http://docs.mongodb.org/ecosystem/tutorial/model-data-for-ruby-on-rails/
When we display our list of stories, we'll need to show the name of the user who posted the story. If we were using a relational database, we could perform a join on users and stores, and get all our objects in a single query. But MongoDB does not support joins and so, at times, requires bit of denormalization. Here, this means caching the 'username' attribute.
Relational purists may be feeling uneasy already, as if we were violating some universal law. But let’s bear in mind that MongoDB collections are not equivalent to relational tables; each serves a unique design objective. A normalized table provides an atomic, isolated chunk of data. A document, however, more closely represents an object as a whole. In the case of a social news site, it can be argued that a username is intrinsic to the story being posted.
114
Then why so many people love MongoDB. I dont get it :|
– Boris D. Teoharov
Sep 15 '13 at 15:47
48
@dudelgrincen it's a paradigm shift from normalization and relational databases. The goal of a NoSQL is to read and write from the database very quickly. With BigData you're going to have scads of application and front end servers with lower numbers on DBs. You're expected to do millions of transactions a second. Offload the heavy lifting from the database and put it onto the application level. If you need deep analysis, you run an integration job that puts your data into an OLAP database. You shouldn't be getting many deep queries from your OLTP dbs anyway.
– Snowburnt
Nov 4 '13 at 1:53
17
@dudelgrincen I should also say that it's not for every project or design. If you have something that works in a SQL type database why change it? If you can't massage your schema to work with noSQL, then don't.
– Snowburnt
Nov 12 '13 at 0:30
8
Migrations and a constantly evolving schemas are also a lot easier to manage on a NoSQL system.
– justin
May 6 '14 at 20:09
11
What if the user has 3.540 posts in the website, and he does change his username in profile? Should every post be updated with the new username?
– Ivo Pereira
Mar 2 '16 at 17:39
|
show 7 more comments
This page on the official mongodb site addresses exactly this question:
http://docs.mongodb.org/ecosystem/tutorial/model-data-for-ruby-on-rails/
When we display our list of stories, we'll need to show the name of the user who posted the story. If we were using a relational database, we could perform a join on users and stores, and get all our objects in a single query. But MongoDB does not support joins and so, at times, requires bit of denormalization. Here, this means caching the 'username' attribute.
Relational purists may be feeling uneasy already, as if we were violating some universal law. But let’s bear in mind that MongoDB collections are not equivalent to relational tables; each serves a unique design objective. A normalized table provides an atomic, isolated chunk of data. A document, however, more closely represents an object as a whole. In the case of a social news site, it can be argued that a username is intrinsic to the story being posted.
114
Then why so many people love MongoDB. I dont get it :|
– Boris D. Teoharov
Sep 15 '13 at 15:47
48
@dudelgrincen it's a paradigm shift from normalization and relational databases. The goal of a NoSQL is to read and write from the database very quickly. With BigData you're going to have scads of application and front end servers with lower numbers on DBs. You're expected to do millions of transactions a second. Offload the heavy lifting from the database and put it onto the application level. If you need deep analysis, you run an integration job that puts your data into an OLAP database. You shouldn't be getting many deep queries from your OLTP dbs anyway.
– Snowburnt
Nov 4 '13 at 1:53
17
@dudelgrincen I should also say that it's not for every project or design. If you have something that works in a SQL type database why change it? If you can't massage your schema to work with noSQL, then don't.
– Snowburnt
Nov 12 '13 at 0:30
8
Migrations and a constantly evolving schemas are also a lot easier to manage on a NoSQL system.
– justin
May 6 '14 at 20:09
11
What if the user has 3.540 posts in the website, and he does change his username in profile? Should every post be updated with the new username?
– Ivo Pereira
Mar 2 '16 at 17:39
|
show 7 more comments
This page on the official mongodb site addresses exactly this question:
http://docs.mongodb.org/ecosystem/tutorial/model-data-for-ruby-on-rails/
When we display our list of stories, we'll need to show the name of the user who posted the story. If we were using a relational database, we could perform a join on users and stores, and get all our objects in a single query. But MongoDB does not support joins and so, at times, requires bit of denormalization. Here, this means caching the 'username' attribute.
Relational purists may be feeling uneasy already, as if we were violating some universal law. But let’s bear in mind that MongoDB collections are not equivalent to relational tables; each serves a unique design objective. A normalized table provides an atomic, isolated chunk of data. A document, however, more closely represents an object as a whole. In the case of a social news site, it can be argued that a username is intrinsic to the story being posted.
This page on the official mongodb site addresses exactly this question:
http://docs.mongodb.org/ecosystem/tutorial/model-data-for-ruby-on-rails/
When we display our list of stories, we'll need to show the name of the user who posted the story. If we were using a relational database, we could perform a join on users and stores, and get all our objects in a single query. But MongoDB does not support joins and so, at times, requires bit of denormalization. Here, this means caching the 'username' attribute.
Relational purists may be feeling uneasy already, as if we were violating some universal law. But let’s bear in mind that MongoDB collections are not equivalent to relational tables; each serves a unique design objective. A normalized table provides an atomic, isolated chunk of data. A document, however, more closely represents an object as a whole. In the case of a social news site, it can be argued that a username is intrinsic to the story being posted.
edited Jul 13 '15 at 17:11
KyleMit
58.3k35241401
58.3k35241401
answered Feb 8 '11 at 20:04
William SteinWilliam Stein
2,0031139
2,0031139
114
Then why so many people love MongoDB. I dont get it :|
– Boris D. Teoharov
Sep 15 '13 at 15:47
48
@dudelgrincen it's a paradigm shift from normalization and relational databases. The goal of a NoSQL is to read and write from the database very quickly. With BigData you're going to have scads of application and front end servers with lower numbers on DBs. You're expected to do millions of transactions a second. Offload the heavy lifting from the database and put it onto the application level. If you need deep analysis, you run an integration job that puts your data into an OLAP database. You shouldn't be getting many deep queries from your OLTP dbs anyway.
– Snowburnt
Nov 4 '13 at 1:53
17
@dudelgrincen I should also say that it's not for every project or design. If you have something that works in a SQL type database why change it? If you can't massage your schema to work with noSQL, then don't.
– Snowburnt
Nov 12 '13 at 0:30
8
Migrations and a constantly evolving schemas are also a lot easier to manage on a NoSQL system.
– justin
May 6 '14 at 20:09
11
What if the user has 3.540 posts in the website, and he does change his username in profile? Should every post be updated with the new username?
– Ivo Pereira
Mar 2 '16 at 17:39
|
show 7 more comments
114
Then why so many people love MongoDB. I dont get it :|
– Boris D. Teoharov
Sep 15 '13 at 15:47
48
@dudelgrincen it's a paradigm shift from normalization and relational databases. The goal of a NoSQL is to read and write from the database very quickly. With BigData you're going to have scads of application and front end servers with lower numbers on DBs. You're expected to do millions of transactions a second. Offload the heavy lifting from the database and put it onto the application level. If you need deep analysis, you run an integration job that puts your data into an OLAP database. You shouldn't be getting many deep queries from your OLTP dbs anyway.
– Snowburnt
Nov 4 '13 at 1:53
17
@dudelgrincen I should also say that it's not for every project or design. If you have something that works in a SQL type database why change it? If you can't massage your schema to work with noSQL, then don't.
– Snowburnt
Nov 12 '13 at 0:30
8
Migrations and a constantly evolving schemas are also a lot easier to manage on a NoSQL system.
– justin
May 6 '14 at 20:09
11
What if the user has 3.540 posts in the website, and he does change his username in profile? Should every post be updated with the new username?
– Ivo Pereira
Mar 2 '16 at 17:39
114
114
Then why so many people love MongoDB. I dont get it :|
– Boris D. Teoharov
Sep 15 '13 at 15:47
Then why so many people love MongoDB. I dont get it :|
– Boris D. Teoharov
Sep 15 '13 at 15:47
48
48
@dudelgrincen it's a paradigm shift from normalization and relational databases. The goal of a NoSQL is to read and write from the database very quickly. With BigData you're going to have scads of application and front end servers with lower numbers on DBs. You're expected to do millions of transactions a second. Offload the heavy lifting from the database and put it onto the application level. If you need deep analysis, you run an integration job that puts your data into an OLAP database. You shouldn't be getting many deep queries from your OLTP dbs anyway.
– Snowburnt
Nov 4 '13 at 1:53
@dudelgrincen it's a paradigm shift from normalization and relational databases. The goal of a NoSQL is to read and write from the database very quickly. With BigData you're going to have scads of application and front end servers with lower numbers on DBs. You're expected to do millions of transactions a second. Offload the heavy lifting from the database and put it onto the application level. If you need deep analysis, you run an integration job that puts your data into an OLAP database. You shouldn't be getting many deep queries from your OLTP dbs anyway.
– Snowburnt
Nov 4 '13 at 1:53
17
17
@dudelgrincen I should also say that it's not for every project or design. If you have something that works in a SQL type database why change it? If you can't massage your schema to work with noSQL, then don't.
– Snowburnt
Nov 12 '13 at 0:30
@dudelgrincen I should also say that it's not for every project or design. If you have something that works in a SQL type database why change it? If you can't massage your schema to work with noSQL, then don't.
– Snowburnt
Nov 12 '13 at 0:30
8
8
Migrations and a constantly evolving schemas are also a lot easier to manage on a NoSQL system.
– justin
May 6 '14 at 20:09
Migrations and a constantly evolving schemas are also a lot easier to manage on a NoSQL system.
– justin
May 6 '14 at 20:09
11
11
What if the user has 3.540 posts in the website, and he does change his username in profile? Should every post be updated with the new username?
– Ivo Pereira
Mar 2 '16 at 17:39
What if the user has 3.540 posts in the website, and he does change his username in profile? Should every post be updated with the new username?
– Ivo Pereira
Mar 2 '16 at 17:39
|
show 7 more comments
We can merge/join all data inside only one collection with a easy function in few lines using the mongodb client console, and now we could be able of perform the desired query.
Below a complete example,
.- Authors:
db.authors.insert([
{
_id: 'a1',
name: { first: 'orlando', last: 'becerra' },
age: 27
},
{
_id: 'a2',
name: { first: 'mayra', last: 'sanchez' },
age: 21
}
]);
.- Categories:
db.categories.insert([
{
_id: 'c1',
name: 'sci-fi'
},
{
_id: 'c2',
name: 'romance'
}
]);
.- Books
db.books.insert([
{
_id: 'b1',
name: 'Groovy Book',
category: 'c1',
authors: ['a1']
},
{
_id: 'b2',
name: 'Java Book',
category: 'c2',
authors: ['a1','a2']
},
]);
.- Book lending
db.lendings.insert([
{
_id: 'l1',
book: 'b1',
date: new Date('01/01/11'),
lendingBy: 'jose'
},
{
_id: 'l2',
book: 'b1',
date: new Date('02/02/12'),
lendingBy: 'maria'
}
]);
.- The magic:
db.books.find().forEach(
function (newBook) {
newBook.category = db.categories.findOne( { "_id": newBook.category } );
newBook.lendings = db.lendings.find( { "book": newBook._id } ).toArray();
newBook.authors = db.authors.find( { "_id": { $in: newBook.authors } } ).toArray();
db.booksReloaded.insert(newBook);
}
);
.- Get the new collection data:
db.booksReloaded.find().pretty()
.- Response :)
{
"_id" : "b1",
"name" : "Groovy Book",
"category" : {
"_id" : "c1",
"name" : "sci-fi"
},
"authors" : [
{
"_id" : "a1",
"name" : {
"first" : "orlando",
"last" : "becerra"
},
"age" : 27
}
],
"lendings" : [
{
"_id" : "l1",
"book" : "b1",
"date" : ISODate("2011-01-01T00:00:00Z"),
"lendingBy" : "jose"
},
{
"_id" : "l2",
"book" : "b1",
"date" : ISODate("2012-02-02T00:00:00Z"),
"lendingBy" : "maria"
}
]
}
{
"_id" : "b2",
"name" : "Java Book",
"category" : {
"_id" : "c2",
"name" : "romance"
},
"authors" : [
{
"_id" : "a1",
"name" : {
"first" : "orlando",
"last" : "becerra"
},
"age" : 27
},
{
"_id" : "a2",
"name" : {
"first" : "mayra",
"last" : "sanchez"
},
"age" : 21
}
],
"lendings" : [ ]
}
I hope this lines can help you.
2
i'm wondering if this same code can be ran using doctrine mongodb?
– abbood
May 30 '14 at 13:46
4
What happens when one of the references objects gets an update? Does that update automatically reflect in the book object? Or does that loop need to run again?
– balupton
Jun 4 '14 at 5:55
9
This is fine as long as your data is small. It is going to bring each book content to your client and then fetch each category, lending and authors one by one. The moment your books are in thousands, this would go really really slow. A better technique probably would be to use aggregation pipeline and output the merged data into a separate collection. Let me get back to it again. I will add that an answer.
– Sandeep Giri
Jun 19 '14 at 15:31
Can you adapt your algorithm to this other example? stackoverflow.com/q/32718079/287948
– Peter Krauss
Sep 22 '15 at 14:13
1
@SandeepGiri how can i do the aggregate pipeline since i have really really intensive data in separated collection need join ??
– Yassine Abdul-Rahman
Oct 7 '15 at 20:16
|
show 3 more comments
We can merge/join all data inside only one collection with a easy function in few lines using the mongodb client console, and now we could be able of perform the desired query.
Below a complete example,
.- Authors:
db.authors.insert([
{
_id: 'a1',
name: { first: 'orlando', last: 'becerra' },
age: 27
},
{
_id: 'a2',
name: { first: 'mayra', last: 'sanchez' },
age: 21
}
]);
.- Categories:
db.categories.insert([
{
_id: 'c1',
name: 'sci-fi'
},
{
_id: 'c2',
name: 'romance'
}
]);
.- Books
db.books.insert([
{
_id: 'b1',
name: 'Groovy Book',
category: 'c1',
authors: ['a1']
},
{
_id: 'b2',
name: 'Java Book',
category: 'c2',
authors: ['a1','a2']
},
]);
.- Book lending
db.lendings.insert([
{
_id: 'l1',
book: 'b1',
date: new Date('01/01/11'),
lendingBy: 'jose'
},
{
_id: 'l2',
book: 'b1',
date: new Date('02/02/12'),
lendingBy: 'maria'
}
]);
.- The magic:
db.books.find().forEach(
function (newBook) {
newBook.category = db.categories.findOne( { "_id": newBook.category } );
newBook.lendings = db.lendings.find( { "book": newBook._id } ).toArray();
newBook.authors = db.authors.find( { "_id": { $in: newBook.authors } } ).toArray();
db.booksReloaded.insert(newBook);
}
);
.- Get the new collection data:
db.booksReloaded.find().pretty()
.- Response :)
{
"_id" : "b1",
"name" : "Groovy Book",
"category" : {
"_id" : "c1",
"name" : "sci-fi"
},
"authors" : [
{
"_id" : "a1",
"name" : {
"first" : "orlando",
"last" : "becerra"
},
"age" : 27
}
],
"lendings" : [
{
"_id" : "l1",
"book" : "b1",
"date" : ISODate("2011-01-01T00:00:00Z"),
"lendingBy" : "jose"
},
{
"_id" : "l2",
"book" : "b1",
"date" : ISODate("2012-02-02T00:00:00Z"),
"lendingBy" : "maria"
}
]
}
{
"_id" : "b2",
"name" : "Java Book",
"category" : {
"_id" : "c2",
"name" : "romance"
},
"authors" : [
{
"_id" : "a1",
"name" : {
"first" : "orlando",
"last" : "becerra"
},
"age" : 27
},
{
"_id" : "a2",
"name" : {
"first" : "mayra",
"last" : "sanchez"
},
"age" : 21
}
],
"lendings" : [ ]
}
I hope this lines can help you.
2
i'm wondering if this same code can be ran using doctrine mongodb?
– abbood
May 30 '14 at 13:46
4
What happens when one of the references objects gets an update? Does that update automatically reflect in the book object? Or does that loop need to run again?
– balupton
Jun 4 '14 at 5:55
9
This is fine as long as your data is small. It is going to bring each book content to your client and then fetch each category, lending and authors one by one. The moment your books are in thousands, this would go really really slow. A better technique probably would be to use aggregation pipeline and output the merged data into a separate collection. Let me get back to it again. I will add that an answer.
– Sandeep Giri
Jun 19 '14 at 15:31
Can you adapt your algorithm to this other example? stackoverflow.com/q/32718079/287948
– Peter Krauss
Sep 22 '15 at 14:13
1
@SandeepGiri how can i do the aggregate pipeline since i have really really intensive data in separated collection need join ??
– Yassine Abdul-Rahman
Oct 7 '15 at 20:16
|
show 3 more comments
We can merge/join all data inside only one collection with a easy function in few lines using the mongodb client console, and now we could be able of perform the desired query.
Below a complete example,
.- Authors:
db.authors.insert([
{
_id: 'a1',
name: { first: 'orlando', last: 'becerra' },
age: 27
},
{
_id: 'a2',
name: { first: 'mayra', last: 'sanchez' },
age: 21
}
]);
.- Categories:
db.categories.insert([
{
_id: 'c1',
name: 'sci-fi'
},
{
_id: 'c2',
name: 'romance'
}
]);
.- Books
db.books.insert([
{
_id: 'b1',
name: 'Groovy Book',
category: 'c1',
authors: ['a1']
},
{
_id: 'b2',
name: 'Java Book',
category: 'c2',
authors: ['a1','a2']
},
]);
.- Book lending
db.lendings.insert([
{
_id: 'l1',
book: 'b1',
date: new Date('01/01/11'),
lendingBy: 'jose'
},
{
_id: 'l2',
book: 'b1',
date: new Date('02/02/12'),
lendingBy: 'maria'
}
]);
.- The magic:
db.books.find().forEach(
function (newBook) {
newBook.category = db.categories.findOne( { "_id": newBook.category } );
newBook.lendings = db.lendings.find( { "book": newBook._id } ).toArray();
newBook.authors = db.authors.find( { "_id": { $in: newBook.authors } } ).toArray();
db.booksReloaded.insert(newBook);
}
);
.- Get the new collection data:
db.booksReloaded.find().pretty()
.- Response :)
{
"_id" : "b1",
"name" : "Groovy Book",
"category" : {
"_id" : "c1",
"name" : "sci-fi"
},
"authors" : [
{
"_id" : "a1",
"name" : {
"first" : "orlando",
"last" : "becerra"
},
"age" : 27
}
],
"lendings" : [
{
"_id" : "l1",
"book" : "b1",
"date" : ISODate("2011-01-01T00:00:00Z"),
"lendingBy" : "jose"
},
{
"_id" : "l2",
"book" : "b1",
"date" : ISODate("2012-02-02T00:00:00Z"),
"lendingBy" : "maria"
}
]
}
{
"_id" : "b2",
"name" : "Java Book",
"category" : {
"_id" : "c2",
"name" : "romance"
},
"authors" : [
{
"_id" : "a1",
"name" : {
"first" : "orlando",
"last" : "becerra"
},
"age" : 27
},
{
"_id" : "a2",
"name" : {
"first" : "mayra",
"last" : "sanchez"
},
"age" : 21
}
],
"lendings" : [ ]
}
I hope this lines can help you.
We can merge/join all data inside only one collection with a easy function in few lines using the mongodb client console, and now we could be able of perform the desired query.
Below a complete example,
.- Authors:
db.authors.insert([
{
_id: 'a1',
name: { first: 'orlando', last: 'becerra' },
age: 27
},
{
_id: 'a2',
name: { first: 'mayra', last: 'sanchez' },
age: 21
}
]);
.- Categories:
db.categories.insert([
{
_id: 'c1',
name: 'sci-fi'
},
{
_id: 'c2',
name: 'romance'
}
]);
.- Books
db.books.insert([
{
_id: 'b1',
name: 'Groovy Book',
category: 'c1',
authors: ['a1']
},
{
_id: 'b2',
name: 'Java Book',
category: 'c2',
authors: ['a1','a2']
},
]);
.- Book lending
db.lendings.insert([
{
_id: 'l1',
book: 'b1',
date: new Date('01/01/11'),
lendingBy: 'jose'
},
{
_id: 'l2',
book: 'b1',
date: new Date('02/02/12'),
lendingBy: 'maria'
}
]);
.- The magic:
db.books.find().forEach(
function (newBook) {
newBook.category = db.categories.findOne( { "_id": newBook.category } );
newBook.lendings = db.lendings.find( { "book": newBook._id } ).toArray();
newBook.authors = db.authors.find( { "_id": { $in: newBook.authors } } ).toArray();
db.booksReloaded.insert(newBook);
}
);
.- Get the new collection data:
db.booksReloaded.find().pretty()
.- Response :)
{
"_id" : "b1",
"name" : "Groovy Book",
"category" : {
"_id" : "c1",
"name" : "sci-fi"
},
"authors" : [
{
"_id" : "a1",
"name" : {
"first" : "orlando",
"last" : "becerra"
},
"age" : 27
}
],
"lendings" : [
{
"_id" : "l1",
"book" : "b1",
"date" : ISODate("2011-01-01T00:00:00Z"),
"lendingBy" : "jose"
},
{
"_id" : "l2",
"book" : "b1",
"date" : ISODate("2012-02-02T00:00:00Z"),
"lendingBy" : "maria"
}
]
}
{
"_id" : "b2",
"name" : "Java Book",
"category" : {
"_id" : "c2",
"name" : "romance"
},
"authors" : [
{
"_id" : "a1",
"name" : {
"first" : "orlando",
"last" : "becerra"
},
"age" : 27
},
{
"_id" : "a2",
"name" : {
"first" : "mayra",
"last" : "sanchez"
},
"age" : 21
}
],
"lendings" : [ ]
}
I hope this lines can help you.
answered Mar 30 '14 at 3:12
Orlando BecerraOrlando Becerra
1,309163
1,309163
2
i'm wondering if this same code can be ran using doctrine mongodb?
– abbood
May 30 '14 at 13:46
4
What happens when one of the references objects gets an update? Does that update automatically reflect in the book object? Or does that loop need to run again?
– balupton
Jun 4 '14 at 5:55
9
This is fine as long as your data is small. It is going to bring each book content to your client and then fetch each category, lending and authors one by one. The moment your books are in thousands, this would go really really slow. A better technique probably would be to use aggregation pipeline and output the merged data into a separate collection. Let me get back to it again. I will add that an answer.
– Sandeep Giri
Jun 19 '14 at 15:31
Can you adapt your algorithm to this other example? stackoverflow.com/q/32718079/287948
– Peter Krauss
Sep 22 '15 at 14:13
1
@SandeepGiri how can i do the aggregate pipeline since i have really really intensive data in separated collection need join ??
– Yassine Abdul-Rahman
Oct 7 '15 at 20:16
|
show 3 more comments
2
i'm wondering if this same code can be ran using doctrine mongodb?
– abbood
May 30 '14 at 13:46
4
What happens when one of the references objects gets an update? Does that update automatically reflect in the book object? Or does that loop need to run again?
– balupton
Jun 4 '14 at 5:55
9
This is fine as long as your data is small. It is going to bring each book content to your client and then fetch each category, lending and authors one by one. The moment your books are in thousands, this would go really really slow. A better technique probably would be to use aggregation pipeline and output the merged data into a separate collection. Let me get back to it again. I will add that an answer.
– Sandeep Giri
Jun 19 '14 at 15:31
Can you adapt your algorithm to this other example? stackoverflow.com/q/32718079/287948
– Peter Krauss
Sep 22 '15 at 14:13
1
@SandeepGiri how can i do the aggregate pipeline since i have really really intensive data in separated collection need join ??
– Yassine Abdul-Rahman
Oct 7 '15 at 20:16
2
2
i'm wondering if this same code can be ran using doctrine mongodb?
– abbood
May 30 '14 at 13:46
i'm wondering if this same code can be ran using doctrine mongodb?
– abbood
May 30 '14 at 13:46
4
4
What happens when one of the references objects gets an update? Does that update automatically reflect in the book object? Or does that loop need to run again?
– balupton
Jun 4 '14 at 5:55
What happens when one of the references objects gets an update? Does that update automatically reflect in the book object? Or does that loop need to run again?
– balupton
Jun 4 '14 at 5:55
9
9
This is fine as long as your data is small. It is going to bring each book content to your client and then fetch each category, lending and authors one by one. The moment your books are in thousands, this would go really really slow. A better technique probably would be to use aggregation pipeline and output the merged data into a separate collection. Let me get back to it again. I will add that an answer.
– Sandeep Giri
Jun 19 '14 at 15:31
This is fine as long as your data is small. It is going to bring each book content to your client and then fetch each category, lending and authors one by one. The moment your books are in thousands, this would go really really slow. A better technique probably would be to use aggregation pipeline and output the merged data into a separate collection. Let me get back to it again. I will add that an answer.
– Sandeep Giri
Jun 19 '14 at 15:31
Can you adapt your algorithm to this other example? stackoverflow.com/q/32718079/287948
– Peter Krauss
Sep 22 '15 at 14:13
Can you adapt your algorithm to this other example? stackoverflow.com/q/32718079/287948
– Peter Krauss
Sep 22 '15 at 14:13
1
1
@SandeepGiri how can i do the aggregate pipeline since i have really really intensive data in separated collection need join ??
– Yassine Abdul-Rahman
Oct 7 '15 at 20:16
@SandeepGiri how can i do the aggregate pipeline since i have really really intensive data in separated collection need join ??
– Yassine Abdul-Rahman
Oct 7 '15 at 20:16
|
show 3 more comments
You have to do it the way you described. MongoDB is a non-relational database and doesn't support joins.
4
Seems wrong performance wise coming from a sql server background, but its maybe not that bad with a document db?
– terjetyl
Jul 15 '10 at 18:20
3
from a sql server background as well, I would appreciate MongoDB taking a 'result set' (with selected returned fields) as input for a new query in one go, much like nested queries in SQL
– Stijn Sanders
Nov 26 '10 at 23:17
1
@terjetyl You have to really plan for it. What fields are you going be presenting on the front end, if it's a limited amount in an individual view then you take those as embedded documents. The key is to not need to do joins. If you want to do deep analysis, you do it after the fact in another database. Run a job that transforms the data into an OLAP cube for optimal performance.
– Snowburnt
Nov 4 '13 at 1:56
3
From mongo 3.2 version left joins are supported.
– Somnath Muluk
Nov 26 '15 at 11:12
add a comment |
You have to do it the way you described. MongoDB is a non-relational database and doesn't support joins.
4
Seems wrong performance wise coming from a sql server background, but its maybe not that bad with a document db?
– terjetyl
Jul 15 '10 at 18:20
3
from a sql server background as well, I would appreciate MongoDB taking a 'result set' (with selected returned fields) as input for a new query in one go, much like nested queries in SQL
– Stijn Sanders
Nov 26 '10 at 23:17
1
@terjetyl You have to really plan for it. What fields are you going be presenting on the front end, if it's a limited amount in an individual view then you take those as embedded documents. The key is to not need to do joins. If you want to do deep analysis, you do it after the fact in another database. Run a job that transforms the data into an OLAP cube for optimal performance.
– Snowburnt
Nov 4 '13 at 1:56
3
From mongo 3.2 version left joins are supported.
– Somnath Muluk
Nov 26 '15 at 11:12
add a comment |
You have to do it the way you described. MongoDB is a non-relational database and doesn't support joins.
You have to do it the way you described. MongoDB is a non-relational database and doesn't support joins.
answered Feb 28 '10 at 11:34
Otto AllmendingerOtto Allmendinger
20.4k45474
20.4k45474
4
Seems wrong performance wise coming from a sql server background, but its maybe not that bad with a document db?
– terjetyl
Jul 15 '10 at 18:20
3
from a sql server background as well, I would appreciate MongoDB taking a 'result set' (with selected returned fields) as input for a new query in one go, much like nested queries in SQL
– Stijn Sanders
Nov 26 '10 at 23:17
1
@terjetyl You have to really plan for it. What fields are you going be presenting on the front end, if it's a limited amount in an individual view then you take those as embedded documents. The key is to not need to do joins. If you want to do deep analysis, you do it after the fact in another database. Run a job that transforms the data into an OLAP cube for optimal performance.
– Snowburnt
Nov 4 '13 at 1:56
3
From mongo 3.2 version left joins are supported.
– Somnath Muluk
Nov 26 '15 at 11:12
add a comment |
4
Seems wrong performance wise coming from a sql server background, but its maybe not that bad with a document db?
– terjetyl
Jul 15 '10 at 18:20
3
from a sql server background as well, I would appreciate MongoDB taking a 'result set' (with selected returned fields) as input for a new query in one go, much like nested queries in SQL
– Stijn Sanders
Nov 26 '10 at 23:17
1
@terjetyl You have to really plan for it. What fields are you going be presenting on the front end, if it's a limited amount in an individual view then you take those as embedded documents. The key is to not need to do joins. If you want to do deep analysis, you do it after the fact in another database. Run a job that transforms the data into an OLAP cube for optimal performance.
– Snowburnt
Nov 4 '13 at 1:56
3
From mongo 3.2 version left joins are supported.
– Somnath Muluk
Nov 26 '15 at 11:12
4
4
Seems wrong performance wise coming from a sql server background, but its maybe not that bad with a document db?
– terjetyl
Jul 15 '10 at 18:20
Seems wrong performance wise coming from a sql server background, but its maybe not that bad with a document db?
– terjetyl
Jul 15 '10 at 18:20
3
3
from a sql server background as well, I would appreciate MongoDB taking a 'result set' (with selected returned fields) as input for a new query in one go, much like nested queries in SQL
– Stijn Sanders
Nov 26 '10 at 23:17
from a sql server background as well, I would appreciate MongoDB taking a 'result set' (with selected returned fields) as input for a new query in one go, much like nested queries in SQL
– Stijn Sanders
Nov 26 '10 at 23:17
1
1
@terjetyl You have to really plan for it. What fields are you going be presenting on the front end, if it's a limited amount in an individual view then you take those as embedded documents. The key is to not need to do joins. If you want to do deep analysis, you do it after the fact in another database. Run a job that transforms the data into an OLAP cube for optimal performance.
– Snowburnt
Nov 4 '13 at 1:56
@terjetyl You have to really plan for it. What fields are you going be presenting on the front end, if it's a limited amount in an individual view then you take those as embedded documents. The key is to not need to do joins. If you want to do deep analysis, you do it after the fact in another database. Run a job that transforms the data into an OLAP cube for optimal performance.
– Snowburnt
Nov 4 '13 at 1:56
3
3
From mongo 3.2 version left joins are supported.
– Somnath Muluk
Nov 26 '15 at 11:12
From mongo 3.2 version left joins are supported.
– Somnath Muluk
Nov 26 '15 at 11:12
add a comment |
Here's an example of a "join" * Actors and Movies collections:
https://github.com/mongodb/cookbook/blob/master/content/patterns/pivot.txt
It makes use of .mapReduce()
method
* join - an alternative to join in document-oriented databases
16
-1, This is NOT joining data from two collections. It is using data from a single collection (actors) pivoting data around. So that things that were keys are now values and values are now keys... very different than a JOIN.
– Evan Teran
May 22 '12 at 17:44
12
This exactly what you have to do, MongoDB is not relational but document oriented. MapReduce allows to play with data with big performance (you can use cluster etc....) but even for simple cases, its very useful !
– Thomas Decaux
Jun 17 '12 at 19:16
Yet the best approach. Very useful thanks
– Singh
Oct 15 '15 at 13:45
add a comment |
Here's an example of a "join" * Actors and Movies collections:
https://github.com/mongodb/cookbook/blob/master/content/patterns/pivot.txt
It makes use of .mapReduce()
method
* join - an alternative to join in document-oriented databases
16
-1, This is NOT joining data from two collections. It is using data from a single collection (actors) pivoting data around. So that things that were keys are now values and values are now keys... very different than a JOIN.
– Evan Teran
May 22 '12 at 17:44
12
This exactly what you have to do, MongoDB is not relational but document oriented. MapReduce allows to play with data with big performance (you can use cluster etc....) but even for simple cases, its very useful !
– Thomas Decaux
Jun 17 '12 at 19:16
Yet the best approach. Very useful thanks
– Singh
Oct 15 '15 at 13:45
add a comment |
Here's an example of a "join" * Actors and Movies collections:
https://github.com/mongodb/cookbook/blob/master/content/patterns/pivot.txt
It makes use of .mapReduce()
method
* join - an alternative to join in document-oriented databases
Here's an example of a "join" * Actors and Movies collections:
https://github.com/mongodb/cookbook/blob/master/content/patterns/pivot.txt
It makes use of .mapReduce()
method
* join - an alternative to join in document-oriented databases
edited Aug 28 '15 at 16:42
JosephSlote
261210
261210
answered Mar 25 '12 at 10:54
antitoxicantitoxic
3,0322942
3,0322942
16
-1, This is NOT joining data from two collections. It is using data from a single collection (actors) pivoting data around. So that things that were keys are now values and values are now keys... very different than a JOIN.
– Evan Teran
May 22 '12 at 17:44
12
This exactly what you have to do, MongoDB is not relational but document oriented. MapReduce allows to play with data with big performance (you can use cluster etc....) but even for simple cases, its very useful !
– Thomas Decaux
Jun 17 '12 at 19:16
Yet the best approach. Very useful thanks
– Singh
Oct 15 '15 at 13:45
add a comment |
16
-1, This is NOT joining data from two collections. It is using data from a single collection (actors) pivoting data around. So that things that were keys are now values and values are now keys... very different than a JOIN.
– Evan Teran
May 22 '12 at 17:44
12
This exactly what you have to do, MongoDB is not relational but document oriented. MapReduce allows to play with data with big performance (you can use cluster etc....) but even for simple cases, its very useful !
– Thomas Decaux
Jun 17 '12 at 19:16
Yet the best approach. Very useful thanks
– Singh
Oct 15 '15 at 13:45
16
16
-1, This is NOT joining data from two collections. It is using data from a single collection (actors) pivoting data around. So that things that were keys are now values and values are now keys... very different than a JOIN.
– Evan Teran
May 22 '12 at 17:44
-1, This is NOT joining data from two collections. It is using data from a single collection (actors) pivoting data around. So that things that were keys are now values and values are now keys... very different than a JOIN.
– Evan Teran
May 22 '12 at 17:44
12
12
This exactly what you have to do, MongoDB is not relational but document oriented. MapReduce allows to play with data with big performance (you can use cluster etc....) but even for simple cases, its very useful !
– Thomas Decaux
Jun 17 '12 at 19:16
This exactly what you have to do, MongoDB is not relational but document oriented. MapReduce allows to play with data with big performance (you can use cluster etc....) but even for simple cases, its very useful !
– Thomas Decaux
Jun 17 '12 at 19:16
Yet the best approach. Very useful thanks
– Singh
Oct 15 '15 at 13:45
Yet the best approach. Very useful thanks
– Singh
Oct 15 '15 at 13:45
add a comment |
As others have pointed out you are trying to create a relational database from none relational database which you really don't want to do but anyways, if you have a case that you have to do this here is a solution you can use. We first do a foreach find on collection A( or in your case users) and then we get each item as an object then we use object property (in your case uid) to lookup in our second collection (in your case comments) if we can find it then we have a match and we can print or do something with it.
Hope this helps you and good luck :)
db.users.find().forEach(
function (object) {
var commonInBoth=db.comments.findOne({ "uid": object.uid} );
if (commonInBoth != null) {
printjson(commonInBoth) ;
printjson(object) ;
}else {
// did not match so we don't care in this case
}
});
1
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Wont this find the item we are currently looping on?
– Skarlinski
Mar 29 '17 at 12:58
This is poor performance wise for large collections.
– Kusaasira Joshua
Feb 7 at 10:08
add a comment |
As others have pointed out you are trying to create a relational database from none relational database which you really don't want to do but anyways, if you have a case that you have to do this here is a solution you can use. We first do a foreach find on collection A( or in your case users) and then we get each item as an object then we use object property (in your case uid) to lookup in our second collection (in your case comments) if we can find it then we have a match and we can print or do something with it.
Hope this helps you and good luck :)
db.users.find().forEach(
function (object) {
var commonInBoth=db.comments.findOne({ "uid": object.uid} );
if (commonInBoth != null) {
printjson(commonInBoth) ;
printjson(object) ;
}else {
// did not match so we don't care in this case
}
});
1
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Wont this find the item we are currently looping on?
– Skarlinski
Mar 29 '17 at 12:58
This is poor performance wise for large collections.
– Kusaasira Joshua
Feb 7 at 10:08
add a comment |
As others have pointed out you are trying to create a relational database from none relational database which you really don't want to do but anyways, if you have a case that you have to do this here is a solution you can use. We first do a foreach find on collection A( or in your case users) and then we get each item as an object then we use object property (in your case uid) to lookup in our second collection (in your case comments) if we can find it then we have a match and we can print or do something with it.
Hope this helps you and good luck :)
db.users.find().forEach(
function (object) {
var commonInBoth=db.comments.findOne({ "uid": object.uid} );
if (commonInBoth != null) {
printjson(commonInBoth) ;
printjson(object) ;
}else {
// did not match so we don't care in this case
}
});
As others have pointed out you are trying to create a relational database from none relational database which you really don't want to do but anyways, if you have a case that you have to do this here is a solution you can use. We first do a foreach find on collection A( or in your case users) and then we get each item as an object then we use object property (in your case uid) to lookup in our second collection (in your case comments) if we can find it then we have a match and we can print or do something with it.
Hope this helps you and good luck :)
db.users.find().forEach(
function (object) {
var commonInBoth=db.comments.findOne({ "uid": object.uid} );
if (commonInBoth != null) {
printjson(commonInBoth) ;
printjson(object) ;
}else {
// did not match so we don't care in this case
}
});
edited Feb 17 '16 at 13:06
VishAl
166116
166116
answered May 4 '15 at 5:49
grepitgrepit
9,39715752
9,39715752
1
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Wont this find the item we are currently looping on?
– Skarlinski
Mar 29 '17 at 12:58
This is poor performance wise for large collections.
– Kusaasira Joshua
Feb 7 at 10:08
add a comment |
1
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Wont this find the item we are currently looping on?
– Skarlinski
Mar 29 '17 at 12:58
This is poor performance wise for large collections.
– Kusaasira Joshua
Feb 7 at 10:08
1
1
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Wont this find the item we are currently looping on?
– Skarlinski
Mar 29 '17 at 12:58
Wont this find the item we are currently looping on?
– Skarlinski
Mar 29 '17 at 12:58
This is poor performance wise for large collections.
– Kusaasira Joshua
Feb 7 at 10:08
This is poor performance wise for large collections.
– Kusaasira Joshua
Feb 7 at 10:08
add a comment |
It depends on what you're trying to do.
You currently have it set up as a normalized database, which is fine, and the way you are doing it is appropriate.
However, there are other ways of doing it.
You could have a posts collection that has imbedded comments for each post with references to the users that you can iteratively query to get. You could store the user's name with the comments, you could store them all in one document.
The thing with NoSQL is it's designed for flexible schemas and very fast reading and writing. In a typical Big Data farm the database is the biggest bottleneck, you have fewer database engines than you do application and front end servers...they're more expensive but more powerful, also hard drive space is very cheap comparatively. Normalization comes from the concept of trying to save space, but it comes with a cost at making your databases perform complicated Joins and verifying the integrity of relationships, performing cascading operations. All of which saves the developers some headaches if they designed the database properly.
With NoSQL, if you accept that redundancy and storage space aren't issues because of their cost (both in processor time required to do updates and hard drive costs to store extra data), denormalizing isn't an issue (for embedded arrays that become hundreds of thousands of items it can be a performance issue, but most of the time that's not a problem). Additionally you'll have several application and front end servers for every database cluster. Have them do the heavy lifting of the joins and let the database servers stick to reading and writing.
TL;DR: What you're doing is fine, and there are other ways of doing it. Check out the mongodb documentation's data model patterns for some great examples. http://docs.mongodb.org/manual/data-modeling/
7
"Normalization comes from the concept of trying to save space" I question this. IMHO normalization comes from the concept of avoiding redundancy. Say you store the name of a user along with a blogpost. What if she marries? In a not normalized model you will have to wade through all posts and change the name. In a normalized model you usually change ONE record.
– DanielKhan
Nov 27 '13 at 13:28
@DanielKhan preventing redundancy and saving space are similar concepts, but on re-analysis I do agree, redundancy is the root cause for this design. I'll reword. Thanks for the note.
– Snowburnt
Nov 27 '13 at 19:15
add a comment |
It depends on what you're trying to do.
You currently have it set up as a normalized database, which is fine, and the way you are doing it is appropriate.
However, there are other ways of doing it.
You could have a posts collection that has imbedded comments for each post with references to the users that you can iteratively query to get. You could store the user's name with the comments, you could store them all in one document.
The thing with NoSQL is it's designed for flexible schemas and very fast reading and writing. In a typical Big Data farm the database is the biggest bottleneck, you have fewer database engines than you do application and front end servers...they're more expensive but more powerful, also hard drive space is very cheap comparatively. Normalization comes from the concept of trying to save space, but it comes with a cost at making your databases perform complicated Joins and verifying the integrity of relationships, performing cascading operations. All of which saves the developers some headaches if they designed the database properly.
With NoSQL, if you accept that redundancy and storage space aren't issues because of their cost (both in processor time required to do updates and hard drive costs to store extra data), denormalizing isn't an issue (for embedded arrays that become hundreds of thousands of items it can be a performance issue, but most of the time that's not a problem). Additionally you'll have several application and front end servers for every database cluster. Have them do the heavy lifting of the joins and let the database servers stick to reading and writing.
TL;DR: What you're doing is fine, and there are other ways of doing it. Check out the mongodb documentation's data model patterns for some great examples. http://docs.mongodb.org/manual/data-modeling/
7
"Normalization comes from the concept of trying to save space" I question this. IMHO normalization comes from the concept of avoiding redundancy. Say you store the name of a user along with a blogpost. What if she marries? In a not normalized model you will have to wade through all posts and change the name. In a normalized model you usually change ONE record.
– DanielKhan
Nov 27 '13 at 13:28
@DanielKhan preventing redundancy and saving space are similar concepts, but on re-analysis I do agree, redundancy is the root cause for this design. I'll reword. Thanks for the note.
– Snowburnt
Nov 27 '13 at 19:15
add a comment |
It depends on what you're trying to do.
You currently have it set up as a normalized database, which is fine, and the way you are doing it is appropriate.
However, there are other ways of doing it.
You could have a posts collection that has imbedded comments for each post with references to the users that you can iteratively query to get. You could store the user's name with the comments, you could store them all in one document.
The thing with NoSQL is it's designed for flexible schemas and very fast reading and writing. In a typical Big Data farm the database is the biggest bottleneck, you have fewer database engines than you do application and front end servers...they're more expensive but more powerful, also hard drive space is very cheap comparatively. Normalization comes from the concept of trying to save space, but it comes with a cost at making your databases perform complicated Joins and verifying the integrity of relationships, performing cascading operations. All of which saves the developers some headaches if they designed the database properly.
With NoSQL, if you accept that redundancy and storage space aren't issues because of their cost (both in processor time required to do updates and hard drive costs to store extra data), denormalizing isn't an issue (for embedded arrays that become hundreds of thousands of items it can be a performance issue, but most of the time that's not a problem). Additionally you'll have several application and front end servers for every database cluster. Have them do the heavy lifting of the joins and let the database servers stick to reading and writing.
TL;DR: What you're doing is fine, and there are other ways of doing it. Check out the mongodb documentation's data model patterns for some great examples. http://docs.mongodb.org/manual/data-modeling/
It depends on what you're trying to do.
You currently have it set up as a normalized database, which is fine, and the way you are doing it is appropriate.
However, there are other ways of doing it.
You could have a posts collection that has imbedded comments for each post with references to the users that you can iteratively query to get. You could store the user's name with the comments, you could store them all in one document.
The thing with NoSQL is it's designed for flexible schemas and very fast reading and writing. In a typical Big Data farm the database is the biggest bottleneck, you have fewer database engines than you do application and front end servers...they're more expensive but more powerful, also hard drive space is very cheap comparatively. Normalization comes from the concept of trying to save space, but it comes with a cost at making your databases perform complicated Joins and verifying the integrity of relationships, performing cascading operations. All of which saves the developers some headaches if they designed the database properly.
With NoSQL, if you accept that redundancy and storage space aren't issues because of their cost (both in processor time required to do updates and hard drive costs to store extra data), denormalizing isn't an issue (for embedded arrays that become hundreds of thousands of items it can be a performance issue, but most of the time that's not a problem). Additionally you'll have several application and front end servers for every database cluster. Have them do the heavy lifting of the joins and let the database servers stick to reading and writing.
TL;DR: What you're doing is fine, and there are other ways of doing it. Check out the mongodb documentation's data model patterns for some great examples. http://docs.mongodb.org/manual/data-modeling/
edited Nov 27 '13 at 19:18
answered Nov 5 '13 at 14:11
SnowburntSnowburnt
3,07542240
3,07542240
7
"Normalization comes from the concept of trying to save space" I question this. IMHO normalization comes from the concept of avoiding redundancy. Say you store the name of a user along with a blogpost. What if she marries? In a not normalized model you will have to wade through all posts and change the name. In a normalized model you usually change ONE record.
– DanielKhan
Nov 27 '13 at 13:28
@DanielKhan preventing redundancy and saving space are similar concepts, but on re-analysis I do agree, redundancy is the root cause for this design. I'll reword. Thanks for the note.
– Snowburnt
Nov 27 '13 at 19:15
add a comment |
7
"Normalization comes from the concept of trying to save space" I question this. IMHO normalization comes from the concept of avoiding redundancy. Say you store the name of a user along with a blogpost. What if she marries? In a not normalized model you will have to wade through all posts and change the name. In a normalized model you usually change ONE record.
– DanielKhan
Nov 27 '13 at 13:28
@DanielKhan preventing redundancy and saving space are similar concepts, but on re-analysis I do agree, redundancy is the root cause for this design. I'll reword. Thanks for the note.
– Snowburnt
Nov 27 '13 at 19:15
7
7
"Normalization comes from the concept of trying to save space" I question this. IMHO normalization comes from the concept of avoiding redundancy. Say you store the name of a user along with a blogpost. What if she marries? In a not normalized model you will have to wade through all posts and change the name. In a normalized model you usually change ONE record.
– DanielKhan
Nov 27 '13 at 13:28
"Normalization comes from the concept of trying to save space" I question this. IMHO normalization comes from the concept of avoiding redundancy. Say you store the name of a user along with a blogpost. What if she marries? In a not normalized model you will have to wade through all posts and change the name. In a normalized model you usually change ONE record.
– DanielKhan
Nov 27 '13 at 13:28
@DanielKhan preventing redundancy and saving space are similar concepts, but on re-analysis I do agree, redundancy is the root cause for this design. I'll reword. Thanks for the note.
– Snowburnt
Nov 27 '13 at 19:15
@DanielKhan preventing redundancy and saving space are similar concepts, but on re-analysis I do agree, redundancy is the root cause for this design. I'll reword. Thanks for the note.
– Snowburnt
Nov 27 '13 at 19:15
add a comment |
There is a specification that a lot of drivers support that's called DBRef.
DBRef is a more formal specification for creating references between documents. DBRefs (generally) include a collection name as well as an object id. Most developers only use DBRefs if the collection can change from one document to the next. If your referenced collection will always be the same, the manual references outlined above are more efficient.
Taken from MongoDB Documentation: Data Models > Data Model Reference >
Database References
add a comment |
There is a specification that a lot of drivers support that's called DBRef.
DBRef is a more formal specification for creating references between documents. DBRefs (generally) include a collection name as well as an object id. Most developers only use DBRefs if the collection can change from one document to the next. If your referenced collection will always be the same, the manual references outlined above are more efficient.
Taken from MongoDB Documentation: Data Models > Data Model Reference >
Database References
add a comment |
There is a specification that a lot of drivers support that's called DBRef.
DBRef is a more formal specification for creating references between documents. DBRefs (generally) include a collection name as well as an object id. Most developers only use DBRefs if the collection can change from one document to the next. If your referenced collection will always be the same, the manual references outlined above are more efficient.
Taken from MongoDB Documentation: Data Models > Data Model Reference >
Database References
There is a specification that a lot of drivers support that's called DBRef.
DBRef is a more formal specification for creating references between documents. DBRefs (generally) include a collection name as well as an object id. Most developers only use DBRefs if the collection can change from one document to the next. If your referenced collection will always be the same, the manual references outlined above are more efficient.
Taken from MongoDB Documentation: Data Models > Data Model Reference >
Database References
edited Nov 23 '15 at 13:13
NDB
493516
493516
answered Mar 1 '10 at 18:27
PickelsPickels
21k20101165
21k20101165
add a comment |
add a comment |
You can join two collection in Mongo by using lookup which is offered in 3.2 version. In your case the query would be
db.comments.aggregate({
$lookup:{
from:"users",
localField:"uid",
foreignField:"uid",
as:"users_comments"
}
})
or you can also join with respect to users then there will be a little change as given below.
db.users.aggregate({
$lookup:{
from:"comments",
localField:"uid",
foreignField:"uid",
as:"users_comments"
}
})
It will work just same as left and right join in SQL.
add a comment |
You can join two collection in Mongo by using lookup which is offered in 3.2 version. In your case the query would be
db.comments.aggregate({
$lookup:{
from:"users",
localField:"uid",
foreignField:"uid",
as:"users_comments"
}
})
or you can also join with respect to users then there will be a little change as given below.
db.users.aggregate({
$lookup:{
from:"comments",
localField:"uid",
foreignField:"uid",
as:"users_comments"
}
})
It will work just same as left and right join in SQL.
add a comment |
You can join two collection in Mongo by using lookup which is offered in 3.2 version. In your case the query would be
db.comments.aggregate({
$lookup:{
from:"users",
localField:"uid",
foreignField:"uid",
as:"users_comments"
}
})
or you can also join with respect to users then there will be a little change as given below.
db.users.aggregate({
$lookup:{
from:"comments",
localField:"uid",
foreignField:"uid",
as:"users_comments"
}
})
It will work just same as left and right join in SQL.
You can join two collection in Mongo by using lookup which is offered in 3.2 version. In your case the query would be
db.comments.aggregate({
$lookup:{
from:"users",
localField:"uid",
foreignField:"uid",
as:"users_comments"
}
})
or you can also join with respect to users then there will be a little change as given below.
db.users.aggregate({
$lookup:{
from:"comments",
localField:"uid",
foreignField:"uid",
as:"users_comments"
}
})
It will work just same as left and right join in SQL.
edited Aug 9 '16 at 13:35
Alex M
2,38071928
2,38071928
answered Aug 9 '16 at 12:29
jarry jaferyjarry jafery
475716
475716
add a comment |
add a comment |
With right combination of $lookup, $project and $match, you can join mutiple tables on multiple parameters. This is because they can be chained multiple times.
Suppose we want to do following (reference)
SELECT S.* FROM LeftTable S
LEFT JOIN RightTable R ON S.ID =R.ID AND S.MID =R.MID WHERE R.TIM >0 AND
S.MOB IS NOT NULL
Step 1: Link all tables
you can $lookup as many tables as you want.
$lookup - one for each table in query
$unwind - because data is denormalised correctly, else wrapped in arrays
Python code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "R"}
])
Step 2: Define all conditionals
$project : define all conditional statements here, plus all the variables you'd like to select.
Python Code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "R"},
# define conditionals + variables
{"$project": {
"midEq": {"$eq": ["$MID", "$R.MID"]},
"ID": 1, "MOB": 1, "MID": 1
}}
])
Step 3: Join all the conditionals
$match - join all conditions using OR or AND etc. There can be multiples of these.
$project: undefine all conditionals
Python Code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "$R"},
# define conditionals + variables
{"$project": {
"midEq": {"$eq": ["$MID", "$R.MID"]},
"ID": 1, "MOB": 1, "MID": 1
}},
# join all conditionals
{"$match": {
"$and": [
{"R.TIM": {"$gt": 0}},
{"MOB": {"$exists": True}},
{"midEq": {"$eq": True}}
]}},
# undefine conditionals
{"$project": {
"midEq": 0
}}
])
Pretty much any combination of tables, conditionals and joins can be done in this manner.
add a comment |
With right combination of $lookup, $project and $match, you can join mutiple tables on multiple parameters. This is because they can be chained multiple times.
Suppose we want to do following (reference)
SELECT S.* FROM LeftTable S
LEFT JOIN RightTable R ON S.ID =R.ID AND S.MID =R.MID WHERE R.TIM >0 AND
S.MOB IS NOT NULL
Step 1: Link all tables
you can $lookup as many tables as you want.
$lookup - one for each table in query
$unwind - because data is denormalised correctly, else wrapped in arrays
Python code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "R"}
])
Step 2: Define all conditionals
$project : define all conditional statements here, plus all the variables you'd like to select.
Python Code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "R"},
# define conditionals + variables
{"$project": {
"midEq": {"$eq": ["$MID", "$R.MID"]},
"ID": 1, "MOB": 1, "MID": 1
}}
])
Step 3: Join all the conditionals
$match - join all conditions using OR or AND etc. There can be multiples of these.
$project: undefine all conditionals
Python Code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "$R"},
# define conditionals + variables
{"$project": {
"midEq": {"$eq": ["$MID", "$R.MID"]},
"ID": 1, "MOB": 1, "MID": 1
}},
# join all conditionals
{"$match": {
"$and": [
{"R.TIM": {"$gt": 0}},
{"MOB": {"$exists": True}},
{"midEq": {"$eq": True}}
]}},
# undefine conditionals
{"$project": {
"midEq": 0
}}
])
Pretty much any combination of tables, conditionals and joins can be done in this manner.
add a comment |
With right combination of $lookup, $project and $match, you can join mutiple tables on multiple parameters. This is because they can be chained multiple times.
Suppose we want to do following (reference)
SELECT S.* FROM LeftTable S
LEFT JOIN RightTable R ON S.ID =R.ID AND S.MID =R.MID WHERE R.TIM >0 AND
S.MOB IS NOT NULL
Step 1: Link all tables
you can $lookup as many tables as you want.
$lookup - one for each table in query
$unwind - because data is denormalised correctly, else wrapped in arrays
Python code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "R"}
])
Step 2: Define all conditionals
$project : define all conditional statements here, plus all the variables you'd like to select.
Python Code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "R"},
# define conditionals + variables
{"$project": {
"midEq": {"$eq": ["$MID", "$R.MID"]},
"ID": 1, "MOB": 1, "MID": 1
}}
])
Step 3: Join all the conditionals
$match - join all conditions using OR or AND etc. There can be multiples of these.
$project: undefine all conditionals
Python Code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "$R"},
# define conditionals + variables
{"$project": {
"midEq": {"$eq": ["$MID", "$R.MID"]},
"ID": 1, "MOB": 1, "MID": 1
}},
# join all conditionals
{"$match": {
"$and": [
{"R.TIM": {"$gt": 0}},
{"MOB": {"$exists": True}},
{"midEq": {"$eq": True}}
]}},
# undefine conditionals
{"$project": {
"midEq": 0
}}
])
Pretty much any combination of tables, conditionals and joins can be done in this manner.
With right combination of $lookup, $project and $match, you can join mutiple tables on multiple parameters. This is because they can be chained multiple times.
Suppose we want to do following (reference)
SELECT S.* FROM LeftTable S
LEFT JOIN RightTable R ON S.ID =R.ID AND S.MID =R.MID WHERE R.TIM >0 AND
S.MOB IS NOT NULL
Step 1: Link all tables
you can $lookup as many tables as you want.
$lookup - one for each table in query
$unwind - because data is denormalised correctly, else wrapped in arrays
Python code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "R"}
])
Step 2: Define all conditionals
$project : define all conditional statements here, plus all the variables you'd like to select.
Python Code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "R"},
# define conditionals + variables
{"$project": {
"midEq": {"$eq": ["$MID", "$R.MID"]},
"ID": 1, "MOB": 1, "MID": 1
}}
])
Step 3: Join all the conditionals
$match - join all conditions using OR or AND etc. There can be multiples of these.
$project: undefine all conditionals
Python Code..
db.LeftTable.aggregate([
# connect all tables
{"$lookup": {
"from": "RightTable",
"localField": "ID",
"foreignField": "ID",
"as": "R"
}},
{"$unwind": "$R"},
# define conditionals + variables
{"$project": {
"midEq": {"$eq": ["$MID", "$R.MID"]},
"ID": 1, "MOB": 1, "MID": 1
}},
# join all conditionals
{"$match": {
"$and": [
{"R.TIM": {"$gt": 0}},
{"MOB": {"$exists": True}},
{"midEq": {"$eq": True}}
]}},
# undefine conditionals
{"$project": {
"midEq": 0
}}
])
Pretty much any combination of tables, conditionals and joins can be done in this manner.
edited May 23 '17 at 12:18
Community♦
11
11
answered Apr 19 '17 at 3:28
Shaurabh BhartiShaurabh Bharti
472513
472513
add a comment |
add a comment |
Before 3.2.6, Mongodb does not support join query as like mysql. below solution which works for you.
db.getCollection('comments').aggregate([
{$match : {pid : 444}},
{$lookup: {from: "users",localField: "uid",foreignField: "uid",as: "userData"}},
])
add a comment |
Before 3.2.6, Mongodb does not support join query as like mysql. below solution which works for you.
db.getCollection('comments').aggregate([
{$match : {pid : 444}},
{$lookup: {from: "users",localField: "uid",foreignField: "uid",as: "userData"}},
])
add a comment |
Before 3.2.6, Mongodb does not support join query as like mysql. below solution which works for you.
db.getCollection('comments').aggregate([
{$match : {pid : 444}},
{$lookup: {from: "users",localField: "uid",foreignField: "uid",as: "userData"}},
])
Before 3.2.6, Mongodb does not support join query as like mysql. below solution which works for you.
db.getCollection('comments').aggregate([
{$match : {pid : 444}},
{$lookup: {from: "users",localField: "uid",foreignField: "uid",as: "userData"}},
])
answered Oct 22 '16 at 4:19
Anish AgarwalAnish Agarwal
1,2511414
1,2511414
add a comment |
add a comment |
You can run SQL queries including join on MongoDB with mongo_fdw from Postgres.
add a comment |
You can run SQL queries including join on MongoDB with mongo_fdw from Postgres.
add a comment |
You can run SQL queries including join on MongoDB with mongo_fdw from Postgres.
You can run SQL queries including join on MongoDB with mongo_fdw from Postgres.
edited Jun 14 '13 at 15:21
answered Oct 18 '12 at 7:02
metdosmetdos
5,461105595
5,461105595
add a comment |
add a comment |
$lookup (aggregation)
Performs a left outer join to an unsharded collection in the same database to filter in documents from the “joined” collection for processing. To each input document, the $lookup stage adds a new array field whose elements are the matching documents from the “joined” collection. The $lookup stage passes these reshaped documents to the next stage.
The $lookup stage has the following syntaxes:
Equality Match
To perform an equality match between a field from the input documents with a field from the documents of the “joined” collection, the $lookup stage has the following syntax:
{
$lookup:
{
from: <collection to join>,
localField: <field from the input documents>,
foreignField: <field from the documents of the "from" collection>,
as: <output array field>
}
}
The operation would correspond to the following pseudo-SQL statement:
SELECT *, <output array field>
FROM collection
WHERE <output array field> IN (SELECT <documents as determined from the pipeline>
FROM <collection to join>
WHERE <pipeline> );
Mongo URL
add a comment |
$lookup (aggregation)
Performs a left outer join to an unsharded collection in the same database to filter in documents from the “joined” collection for processing. To each input document, the $lookup stage adds a new array field whose elements are the matching documents from the “joined” collection. The $lookup stage passes these reshaped documents to the next stage.
The $lookup stage has the following syntaxes:
Equality Match
To perform an equality match between a field from the input documents with a field from the documents of the “joined” collection, the $lookup stage has the following syntax:
{
$lookup:
{
from: <collection to join>,
localField: <field from the input documents>,
foreignField: <field from the documents of the "from" collection>,
as: <output array field>
}
}
The operation would correspond to the following pseudo-SQL statement:
SELECT *, <output array field>
FROM collection
WHERE <output array field> IN (SELECT <documents as determined from the pipeline>
FROM <collection to join>
WHERE <pipeline> );
Mongo URL
add a comment |
$lookup (aggregation)
Performs a left outer join to an unsharded collection in the same database to filter in documents from the “joined” collection for processing. To each input document, the $lookup stage adds a new array field whose elements are the matching documents from the “joined” collection. The $lookup stage passes these reshaped documents to the next stage.
The $lookup stage has the following syntaxes:
Equality Match
To perform an equality match between a field from the input documents with a field from the documents of the “joined” collection, the $lookup stage has the following syntax:
{
$lookup:
{
from: <collection to join>,
localField: <field from the input documents>,
foreignField: <field from the documents of the "from" collection>,
as: <output array field>
}
}
The operation would correspond to the following pseudo-SQL statement:
SELECT *, <output array field>
FROM collection
WHERE <output array field> IN (SELECT <documents as determined from the pipeline>
FROM <collection to join>
WHERE <pipeline> );
Mongo URL
$lookup (aggregation)
Performs a left outer join to an unsharded collection in the same database to filter in documents from the “joined” collection for processing. To each input document, the $lookup stage adds a new array field whose elements are the matching documents from the “joined” collection. The $lookup stage passes these reshaped documents to the next stage.
The $lookup stage has the following syntaxes:
Equality Match
To perform an equality match between a field from the input documents with a field from the documents of the “joined” collection, the $lookup stage has the following syntax:
{
$lookup:
{
from: <collection to join>,
localField: <field from the input documents>,
foreignField: <field from the documents of the "from" collection>,
as: <output array field>
}
}
The operation would correspond to the following pseudo-SQL statement:
SELECT *, <output array field>
FROM collection
WHERE <output array field> IN (SELECT <documents as determined from the pipeline>
FROM <collection to join>
WHERE <pipeline> );
Mongo URL
answered Mar 27 '18 at 12:19
GoutamSGoutamS
1,356918
1,356918
add a comment |
add a comment |
MongoDB does not allow joins, but you can use plugins to handle that. Check the mongo-join plugin. It's the best and I have already used it. You can install it using npm directly like this npm install mongo-join
. You can check out the full documentation with examples.
(++) really helpful tool when we need to join (N) collections
(--) we can apply conditions just on the top level of the query
Example
var Join = require('mongo-join').Join, mongodb = require('mongodb'), Db = mongodb.Db, Server = mongodb.Server;
db.open(function (err, Database) {
Database.collection('Appoint', function (err, Appoints) {
/* we can put conditions just on the top level */
Appoints.find({_id_Doctor: id_doctor ,full_date :{ $gte: start_date },
full_date :{ $lte: end_date }}, function (err, cursor) {
var join = new Join(Database).on({
field: '_id_Doctor', // <- field in Appoints document
to: '_id', // <- field in User doc. treated as ObjectID automatically.
from: 'User' // <- collection name for User doc
}).on({
field: '_id_Patient', // <- field in Appoints doc
to: '_id', // <- field in User doc. treated as ObjectID automatically.
from: 'User' // <- collection name for User doc
})
join.toArray(cursor, function (err, joinedDocs) {
/* do what ever you want here */
/* you can fetch the table and apply your own conditions */
.....
.....
.....
resp.status(200);
resp.json({
"status": 200,
"message": "success",
"Appoints_Range": joinedDocs,
});
return resp;
});
});
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Indeed I upvoted due to is a solution
– Maga
Oct 29 '15 at 10:21
add a comment |
MongoDB does not allow joins, but you can use plugins to handle that. Check the mongo-join plugin. It's the best and I have already used it. You can install it using npm directly like this npm install mongo-join
. You can check out the full documentation with examples.
(++) really helpful tool when we need to join (N) collections
(--) we can apply conditions just on the top level of the query
Example
var Join = require('mongo-join').Join, mongodb = require('mongodb'), Db = mongodb.Db, Server = mongodb.Server;
db.open(function (err, Database) {
Database.collection('Appoint', function (err, Appoints) {
/* we can put conditions just on the top level */
Appoints.find({_id_Doctor: id_doctor ,full_date :{ $gte: start_date },
full_date :{ $lte: end_date }}, function (err, cursor) {
var join = new Join(Database).on({
field: '_id_Doctor', // <- field in Appoints document
to: '_id', // <- field in User doc. treated as ObjectID automatically.
from: 'User' // <- collection name for User doc
}).on({
field: '_id_Patient', // <- field in Appoints doc
to: '_id', // <- field in User doc. treated as ObjectID automatically.
from: 'User' // <- collection name for User doc
})
join.toArray(cursor, function (err, joinedDocs) {
/* do what ever you want here */
/* you can fetch the table and apply your own conditions */
.....
.....
.....
resp.status(200);
resp.json({
"status": 200,
"message": "success",
"Appoints_Range": joinedDocs,
});
return resp;
});
});
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Indeed I upvoted due to is a solution
– Maga
Oct 29 '15 at 10:21
add a comment |
MongoDB does not allow joins, but you can use plugins to handle that. Check the mongo-join plugin. It's the best and I have already used it. You can install it using npm directly like this npm install mongo-join
. You can check out the full documentation with examples.
(++) really helpful tool when we need to join (N) collections
(--) we can apply conditions just on the top level of the query
Example
var Join = require('mongo-join').Join, mongodb = require('mongodb'), Db = mongodb.Db, Server = mongodb.Server;
db.open(function (err, Database) {
Database.collection('Appoint', function (err, Appoints) {
/* we can put conditions just on the top level */
Appoints.find({_id_Doctor: id_doctor ,full_date :{ $gte: start_date },
full_date :{ $lte: end_date }}, function (err, cursor) {
var join = new Join(Database).on({
field: '_id_Doctor', // <- field in Appoints document
to: '_id', // <- field in User doc. treated as ObjectID automatically.
from: 'User' // <- collection name for User doc
}).on({
field: '_id_Patient', // <- field in Appoints doc
to: '_id', // <- field in User doc. treated as ObjectID automatically.
from: 'User' // <- collection name for User doc
})
join.toArray(cursor, function (err, joinedDocs) {
/* do what ever you want here */
/* you can fetch the table and apply your own conditions */
.....
.....
.....
resp.status(200);
resp.json({
"status": 200,
"message": "success",
"Appoints_Range": joinedDocs,
});
return resp;
});
});
MongoDB does not allow joins, but you can use plugins to handle that. Check the mongo-join plugin. It's the best and I have already used it. You can install it using npm directly like this npm install mongo-join
. You can check out the full documentation with examples.
(++) really helpful tool when we need to join (N) collections
(--) we can apply conditions just on the top level of the query
Example
var Join = require('mongo-join').Join, mongodb = require('mongodb'), Db = mongodb.Db, Server = mongodb.Server;
db.open(function (err, Database) {
Database.collection('Appoint', function (err, Appoints) {
/* we can put conditions just on the top level */
Appoints.find({_id_Doctor: id_doctor ,full_date :{ $gte: start_date },
full_date :{ $lte: end_date }}, function (err, cursor) {
var join = new Join(Database).on({
field: '_id_Doctor', // <- field in Appoints document
to: '_id', // <- field in User doc. treated as ObjectID automatically.
from: 'User' // <- collection name for User doc
}).on({
field: '_id_Patient', // <- field in Appoints doc
to: '_id', // <- field in User doc. treated as ObjectID automatically.
from: 'User' // <- collection name for User doc
})
join.toArray(cursor, function (err, joinedDocs) {
/* do what ever you want here */
/* you can fetch the table and apply your own conditions */
.....
.....
.....
resp.status(200);
resp.json({
"status": 200,
"message": "success",
"Appoints_Range": joinedDocs,
});
return resp;
});
});
edited Oct 15 '15 at 21:15
Michael Mior
21.5k66292
21.5k66292
answered Apr 14 '15 at 15:33
Amine_DevAmine_Dev
427311
427311
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Indeed I upvoted due to is a solution
– Maga
Oct 29 '15 at 10:21
add a comment |
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Indeed I upvoted due to is a solution
– Maga
Oct 29 '15 at 10:21
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Don't see any reason for this to be downvoted.
– Michael Cole
Sep 21 '15 at 14:59
Indeed I upvoted due to is a solution
– Maga
Oct 29 '15 at 10:21
Indeed I upvoted due to is a solution
– Maga
Oct 29 '15 at 10:21
add a comment |
playORM can do it for you using S-SQL(Scalable SQL) which just adds partitioning such that you can do joins within partitions.
add a comment |
playORM can do it for you using S-SQL(Scalable SQL) which just adds partitioning such that you can do joins within partitions.
add a comment |
playORM can do it for you using S-SQL(Scalable SQL) which just adds partitioning such that you can do joins within partitions.
playORM can do it for you using S-SQL(Scalable SQL) which just adds partitioning such that you can do joins within partitions.
answered Sep 7 '12 at 17:30
Dean HillerDean Hiller
10.3k1586148
10.3k1586148
add a comment |
add a comment |
You can do it using the aggregation pipeline, but it's a pain to write it yourself.
You can use mongo-join-query
to create the aggregation pipeline automatically from your query.
This is how your query would look like:
const mongoose = require("mongoose");
const joinQuery = require("mongo-join-query");
joinQuery(
mongoose.models.Comment,
{
find: { pid:444 },
populate: ["uid"]
},
(err, res) => (err ? console.log("Error:", err) : console.log("Success:", res.results))
);
Your result would have the user object in the uid
field and you can link as many levels deep as you want. You can populate the reference to the user, which makes reference to a Team, which makes reference to something else, etc..
Disclaimer: I wrote mongo-join-query
to tackle this exact problem.
add a comment |
You can do it using the aggregation pipeline, but it's a pain to write it yourself.
You can use mongo-join-query
to create the aggregation pipeline automatically from your query.
This is how your query would look like:
const mongoose = require("mongoose");
const joinQuery = require("mongo-join-query");
joinQuery(
mongoose.models.Comment,
{
find: { pid:444 },
populate: ["uid"]
},
(err, res) => (err ? console.log("Error:", err) : console.log("Success:", res.results))
);
Your result would have the user object in the uid
field and you can link as many levels deep as you want. You can populate the reference to the user, which makes reference to a Team, which makes reference to something else, etc..
Disclaimer: I wrote mongo-join-query
to tackle this exact problem.
add a comment |
You can do it using the aggregation pipeline, but it's a pain to write it yourself.
You can use mongo-join-query
to create the aggregation pipeline automatically from your query.
This is how your query would look like:
const mongoose = require("mongoose");
const joinQuery = require("mongo-join-query");
joinQuery(
mongoose.models.Comment,
{
find: { pid:444 },
populate: ["uid"]
},
(err, res) => (err ? console.log("Error:", err) : console.log("Success:", res.results))
);
Your result would have the user object in the uid
field and you can link as many levels deep as you want. You can populate the reference to the user, which makes reference to a Team, which makes reference to something else, etc..
Disclaimer: I wrote mongo-join-query
to tackle this exact problem.
You can do it using the aggregation pipeline, but it's a pain to write it yourself.
You can use mongo-join-query
to create the aggregation pipeline automatically from your query.
This is how your query would look like:
const mongoose = require("mongoose");
const joinQuery = require("mongo-join-query");
joinQuery(
mongoose.models.Comment,
{
find: { pid:444 },
populate: ["uid"]
},
(err, res) => (err ? console.log("Error:", err) : console.log("Success:", res.results))
);
Your result would have the user object in the uid
field and you can link as many levels deep as you want. You can populate the reference to the user, which makes reference to a Team, which makes reference to something else, etc..
Disclaimer: I wrote mongo-join-query
to tackle this exact problem.
answered Dec 13 '17 at 14:28
Marcelo LazaroniMarcelo Lazaroni
3,70811923
3,70811923
add a comment |
add a comment |
Nope, it doesn't seem like you're doing it wrong. MongoDB joins are "client side". Pretty much like you said:
At the moment, I am first getting the comments which match my criteria, then figuring out all the uid's in that result set, getting the user objects, and merging them with the comment's results. Seems like I am doing it wrong.
1) Select from the collection you're interested in.
2) From that collection pull out ID's you need
3) Select from other collections
4) Decorate your original results.
It's not a "real" join, but it's actually alot more useful than a SQL join because you don't have to deal with duplicate rows for "many" sided joins, instead your decorating the originally selected set.
There is alot of nonsense and FUD on this page. Turns out 5 years later MongoDB is still a thing.
'you don't have to deal with duplicate rows for "many" sided joins' - no idea what you mean by this. Can you clarify?
– Mark Amery
Sep 20 '15 at 20:23
Downvote, really?
– Michael Cole
Sep 21 '15 at 14:46
1
@MarkAmery, sure. In SQL a n-n relationship will return duplicate rows. E.g. Friends. If Bob is friends with Mary and Jane, you'll get 2 rows for Bob: Bob,Mary and Bob,Jane. 2 Bobs is a lie, there is only one Bob. With client-side joins you can start with Bob and decorate how you like: Bob, "Mary and Jane". SQL let's you do this with subqueries, but that's doing work on the db server that could be done on the client.
– Michael Cole
Sep 21 '15 at 14:51
add a comment |
Nope, it doesn't seem like you're doing it wrong. MongoDB joins are "client side". Pretty much like you said:
At the moment, I am first getting the comments which match my criteria, then figuring out all the uid's in that result set, getting the user objects, and merging them with the comment's results. Seems like I am doing it wrong.
1) Select from the collection you're interested in.
2) From that collection pull out ID's you need
3) Select from other collections
4) Decorate your original results.
It's not a "real" join, but it's actually alot more useful than a SQL join because you don't have to deal with duplicate rows for "many" sided joins, instead your decorating the originally selected set.
There is alot of nonsense and FUD on this page. Turns out 5 years later MongoDB is still a thing.
'you don't have to deal with duplicate rows for "many" sided joins' - no idea what you mean by this. Can you clarify?
– Mark Amery
Sep 20 '15 at 20:23
Downvote, really?
– Michael Cole
Sep 21 '15 at 14:46
1
@MarkAmery, sure. In SQL a n-n relationship will return duplicate rows. E.g. Friends. If Bob is friends with Mary and Jane, you'll get 2 rows for Bob: Bob,Mary and Bob,Jane. 2 Bobs is a lie, there is only one Bob. With client-side joins you can start with Bob and decorate how you like: Bob, "Mary and Jane". SQL let's you do this with subqueries, but that's doing work on the db server that could be done on the client.
– Michael Cole
Sep 21 '15 at 14:51
add a comment |
Nope, it doesn't seem like you're doing it wrong. MongoDB joins are "client side". Pretty much like you said:
At the moment, I am first getting the comments which match my criteria, then figuring out all the uid's in that result set, getting the user objects, and merging them with the comment's results. Seems like I am doing it wrong.
1) Select from the collection you're interested in.
2) From that collection pull out ID's you need
3) Select from other collections
4) Decorate your original results.
It's not a "real" join, but it's actually alot more useful than a SQL join because you don't have to deal with duplicate rows for "many" sided joins, instead your decorating the originally selected set.
There is alot of nonsense and FUD on this page. Turns out 5 years later MongoDB is still a thing.
Nope, it doesn't seem like you're doing it wrong. MongoDB joins are "client side". Pretty much like you said:
At the moment, I am first getting the comments which match my criteria, then figuring out all the uid's in that result set, getting the user objects, and merging them with the comment's results. Seems like I am doing it wrong.
1) Select from the collection you're interested in.
2) From that collection pull out ID's you need
3) Select from other collections
4) Decorate your original results.
It's not a "real" join, but it's actually alot more useful than a SQL join because you don't have to deal with duplicate rows for "many" sided joins, instead your decorating the originally selected set.
There is alot of nonsense and FUD on this page. Turns out 5 years later MongoDB is still a thing.
answered Sep 4 '15 at 15:23
Michael ColeMichael Cole
9,30334255
9,30334255
'you don't have to deal with duplicate rows for "many" sided joins' - no idea what you mean by this. Can you clarify?
– Mark Amery
Sep 20 '15 at 20:23
Downvote, really?
– Michael Cole
Sep 21 '15 at 14:46
1
@MarkAmery, sure. In SQL a n-n relationship will return duplicate rows. E.g. Friends. If Bob is friends with Mary and Jane, you'll get 2 rows for Bob: Bob,Mary and Bob,Jane. 2 Bobs is a lie, there is only one Bob. With client-side joins you can start with Bob and decorate how you like: Bob, "Mary and Jane". SQL let's you do this with subqueries, but that's doing work on the db server that could be done on the client.
– Michael Cole
Sep 21 '15 at 14:51
add a comment |
'you don't have to deal with duplicate rows for "many" sided joins' - no idea what you mean by this. Can you clarify?
– Mark Amery
Sep 20 '15 at 20:23
Downvote, really?
– Michael Cole
Sep 21 '15 at 14:46
1
@MarkAmery, sure. In SQL a n-n relationship will return duplicate rows. E.g. Friends. If Bob is friends with Mary and Jane, you'll get 2 rows for Bob: Bob,Mary and Bob,Jane. 2 Bobs is a lie, there is only one Bob. With client-side joins you can start with Bob and decorate how you like: Bob, "Mary and Jane". SQL let's you do this with subqueries, but that's doing work on the db server that could be done on the client.
– Michael Cole
Sep 21 '15 at 14:51
'you don't have to deal with duplicate rows for "many" sided joins' - no idea what you mean by this. Can you clarify?
– Mark Amery
Sep 20 '15 at 20:23
'you don't have to deal with duplicate rows for "many" sided joins' - no idea what you mean by this. Can you clarify?
– Mark Amery
Sep 20 '15 at 20:23
Downvote, really?
– Michael Cole
Sep 21 '15 at 14:46
Downvote, really?
– Michael Cole
Sep 21 '15 at 14:46
1
1
@MarkAmery, sure. In SQL a n-n relationship will return duplicate rows. E.g. Friends. If Bob is friends with Mary and Jane, you'll get 2 rows for Bob: Bob,Mary and Bob,Jane. 2 Bobs is a lie, there is only one Bob. With client-side joins you can start with Bob and decorate how you like: Bob, "Mary and Jane". SQL let's you do this with subqueries, but that's doing work on the db server that could be done on the client.
– Michael Cole
Sep 21 '15 at 14:51
@MarkAmery, sure. In SQL a n-n relationship will return duplicate rows. E.g. Friends. If Bob is friends with Mary and Jane, you'll get 2 rows for Bob: Bob,Mary and Bob,Jane. 2 Bobs is a lie, there is only one Bob. With client-side joins you can start with Bob and decorate how you like: Bob, "Mary and Jane". SQL let's you do this with subqueries, but that's doing work on the db server that could be done on the client.
– Michael Cole
Sep 21 '15 at 14:51
add a comment |
I think, if You need normalized data tables - You need to try some other database solutions.
But I've foun that sollution for MOngo on Git
By the way, in inserts code - it has movie's name, but noi movie's ID.
Problem
You have a collection of Actors with an array of the Movies they've done.
You want to generate a collection of Movies with an array of Actors in each.
Some sample data
db.actors.insert( { actor: "Richard Gere", movies: ['Pretty Woman', 'Runaway Bride', 'Chicago'] });
db.actors.insert( { actor: "Julia Roberts", movies: ['Pretty Woman', 'Runaway Bride', 'Erin Brockovich'] });
Solution
We need to loop through each movie in the Actor document and emit each Movie individually.
The catch here is in the reduce phase. We cannot emit an array from the reduce phase, so we must build an Actors array inside of the "value" document that is returned.
The code
map = function() {
for(var i in this.movies){
key = { movie: this.movies[i] };
value = { actors: [ this.actor ] };
emit(key, value);
}
}
reduce = function(key, values) {
actor_list = { actors: };
for(var i in values) {
actor_list.actors = values[i].actors.concat(actor_list.actors);
}
return actor_list;
}
Notice how actor_list is actually a javascript object that contains an array. Also notice that map emits the same structure.
Run the following to execute the map / reduce, output it to the "pivot" collection and print the result:
printjson(db.actors.mapReduce(map, reduce, "pivot"));
db.pivot.find().forEach(printjson);
Here is the sample output, note that "Pretty Woman" and "Runaway Bride" have both "Richard Gere" and "Julia Roberts".
{ "_id" : { "movie" : "Chicago" }, "value" : { "actors" : [ "Richard Gere" ] } }
{ "_id" : { "movie" : "Erin Brockovich" }, "value" : { "actors" : [ "Julia Roberts" ] } }
{ "_id" : { "movie" : "Pretty Woman" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
{ "_id" : { "movie" : "Runaway Bride" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
Note that most of the content of this answer (i.e. the bit that's in comprehensible English) is copied from the MongoDB cookbook at the GitHub link the answerer provided.
– Mark Amery
Oct 5 '15 at 14:23
add a comment |
I think, if You need normalized data tables - You need to try some other database solutions.
But I've foun that sollution for MOngo on Git
By the way, in inserts code - it has movie's name, but noi movie's ID.
Problem
You have a collection of Actors with an array of the Movies they've done.
You want to generate a collection of Movies with an array of Actors in each.
Some sample data
db.actors.insert( { actor: "Richard Gere", movies: ['Pretty Woman', 'Runaway Bride', 'Chicago'] });
db.actors.insert( { actor: "Julia Roberts", movies: ['Pretty Woman', 'Runaway Bride', 'Erin Brockovich'] });
Solution
We need to loop through each movie in the Actor document and emit each Movie individually.
The catch here is in the reduce phase. We cannot emit an array from the reduce phase, so we must build an Actors array inside of the "value" document that is returned.
The code
map = function() {
for(var i in this.movies){
key = { movie: this.movies[i] };
value = { actors: [ this.actor ] };
emit(key, value);
}
}
reduce = function(key, values) {
actor_list = { actors: };
for(var i in values) {
actor_list.actors = values[i].actors.concat(actor_list.actors);
}
return actor_list;
}
Notice how actor_list is actually a javascript object that contains an array. Also notice that map emits the same structure.
Run the following to execute the map / reduce, output it to the "pivot" collection and print the result:
printjson(db.actors.mapReduce(map, reduce, "pivot"));
db.pivot.find().forEach(printjson);
Here is the sample output, note that "Pretty Woman" and "Runaway Bride" have both "Richard Gere" and "Julia Roberts".
{ "_id" : { "movie" : "Chicago" }, "value" : { "actors" : [ "Richard Gere" ] } }
{ "_id" : { "movie" : "Erin Brockovich" }, "value" : { "actors" : [ "Julia Roberts" ] } }
{ "_id" : { "movie" : "Pretty Woman" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
{ "_id" : { "movie" : "Runaway Bride" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
Note that most of the content of this answer (i.e. the bit that's in comprehensible English) is copied from the MongoDB cookbook at the GitHub link the answerer provided.
– Mark Amery
Oct 5 '15 at 14:23
add a comment |
I think, if You need normalized data tables - You need to try some other database solutions.
But I've foun that sollution for MOngo on Git
By the way, in inserts code - it has movie's name, but noi movie's ID.
Problem
You have a collection of Actors with an array of the Movies they've done.
You want to generate a collection of Movies with an array of Actors in each.
Some sample data
db.actors.insert( { actor: "Richard Gere", movies: ['Pretty Woman', 'Runaway Bride', 'Chicago'] });
db.actors.insert( { actor: "Julia Roberts", movies: ['Pretty Woman', 'Runaway Bride', 'Erin Brockovich'] });
Solution
We need to loop through each movie in the Actor document and emit each Movie individually.
The catch here is in the reduce phase. We cannot emit an array from the reduce phase, so we must build an Actors array inside of the "value" document that is returned.
The code
map = function() {
for(var i in this.movies){
key = { movie: this.movies[i] };
value = { actors: [ this.actor ] };
emit(key, value);
}
}
reduce = function(key, values) {
actor_list = { actors: };
for(var i in values) {
actor_list.actors = values[i].actors.concat(actor_list.actors);
}
return actor_list;
}
Notice how actor_list is actually a javascript object that contains an array. Also notice that map emits the same structure.
Run the following to execute the map / reduce, output it to the "pivot" collection and print the result:
printjson(db.actors.mapReduce(map, reduce, "pivot"));
db.pivot.find().forEach(printjson);
Here is the sample output, note that "Pretty Woman" and "Runaway Bride" have both "Richard Gere" and "Julia Roberts".
{ "_id" : { "movie" : "Chicago" }, "value" : { "actors" : [ "Richard Gere" ] } }
{ "_id" : { "movie" : "Erin Brockovich" }, "value" : { "actors" : [ "Julia Roberts" ] } }
{ "_id" : { "movie" : "Pretty Woman" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
{ "_id" : { "movie" : "Runaway Bride" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
I think, if You need normalized data tables - You need to try some other database solutions.
But I've foun that sollution for MOngo on Git
By the way, in inserts code - it has movie's name, but noi movie's ID.
Problem
You have a collection of Actors with an array of the Movies they've done.
You want to generate a collection of Movies with an array of Actors in each.
Some sample data
db.actors.insert( { actor: "Richard Gere", movies: ['Pretty Woman', 'Runaway Bride', 'Chicago'] });
db.actors.insert( { actor: "Julia Roberts", movies: ['Pretty Woman', 'Runaway Bride', 'Erin Brockovich'] });
Solution
We need to loop through each movie in the Actor document and emit each Movie individually.
The catch here is in the reduce phase. We cannot emit an array from the reduce phase, so we must build an Actors array inside of the "value" document that is returned.
The code
map = function() {
for(var i in this.movies){
key = { movie: this.movies[i] };
value = { actors: [ this.actor ] };
emit(key, value);
}
}
reduce = function(key, values) {
actor_list = { actors: };
for(var i in values) {
actor_list.actors = values[i].actors.concat(actor_list.actors);
}
return actor_list;
}
Notice how actor_list is actually a javascript object that contains an array. Also notice that map emits the same structure.
Run the following to execute the map / reduce, output it to the "pivot" collection and print the result:
printjson(db.actors.mapReduce(map, reduce, "pivot"));
db.pivot.find().forEach(printjson);
Here is the sample output, note that "Pretty Woman" and "Runaway Bride" have both "Richard Gere" and "Julia Roberts".
{ "_id" : { "movie" : "Chicago" }, "value" : { "actors" : [ "Richard Gere" ] } }
{ "_id" : { "movie" : "Erin Brockovich" }, "value" : { "actors" : [ "Julia Roberts" ] } }
{ "_id" : { "movie" : "Pretty Woman" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
{ "_id" : { "movie" : "Runaway Bride" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
answered Sep 23 '15 at 6:57
Max SherbakovMax Sherbakov
33927
33927
Note that most of the content of this answer (i.e. the bit that's in comprehensible English) is copied from the MongoDB cookbook at the GitHub link the answerer provided.
– Mark Amery
Oct 5 '15 at 14:23
add a comment |
Note that most of the content of this answer (i.e. the bit that's in comprehensible English) is copied from the MongoDB cookbook at the GitHub link the answerer provided.
– Mark Amery
Oct 5 '15 at 14:23
Note that most of the content of this answer (i.e. the bit that's in comprehensible English) is copied from the MongoDB cookbook at the GitHub link the answerer provided.
– Mark Amery
Oct 5 '15 at 14:23
Note that most of the content of this answer (i.e. the bit that's in comprehensible English) is copied from the MongoDB cookbook at the GitHub link the answerer provided.
– Mark Amery
Oct 5 '15 at 14:23
add a comment |
We can merge two collection by using mongoDB sub query. Here is example,
Commentss--
`db.commentss.insert([
{ uid:12345, pid:444, comment:"blah" },
{ uid:12345, pid:888, comment:"asdf" },
{ uid:99999, pid:444, comment:"qwer" }])`
Userss--
db.userss.insert([
{ uid:12345, name:"john" },
{ uid:99999, name:"mia" }])
MongoDB sub query for JOIN--
`db.commentss.find().forEach(
function (newComments) {
newComments.userss = db.userss.find( { "uid": newComments.uid } ).toArray();
db.newCommentUsers.insert(newComments);
}
);`
Get result from newly generated Collection--
db.newCommentUsers.find().pretty()
Result--
`{
"_id" : ObjectId("5511236e29709afa03f226ef"),
"uid" : 12345,
"pid" : 444,
"comment" : "blah",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f2"),
"uid" : 12345,
"name" : "john"
}
]
}
{
"_id" : ObjectId("5511236e29709afa03f226f0"),
"uid" : 12345,
"pid" : 888,
"comment" : "asdf",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f2"),
"uid" : 12345,
"name" : "john"
}
]
}
{
"_id" : ObjectId("5511236e29709afa03f226f1"),
"uid" : 99999,
"pid" : 444,
"comment" : "qwer",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f3"),
"uid" : 99999,
"name" : "mia"
}
]
}`
Hope so this will help.
7
Why did you basically copy this nearly identical, one-year-old answer? stackoverflow.com/a/22739813/4186945
– hackel
May 5 '15 at 20:45
add a comment |
We can merge two collection by using mongoDB sub query. Here is example,
Commentss--
`db.commentss.insert([
{ uid:12345, pid:444, comment:"blah" },
{ uid:12345, pid:888, comment:"asdf" },
{ uid:99999, pid:444, comment:"qwer" }])`
Userss--
db.userss.insert([
{ uid:12345, name:"john" },
{ uid:99999, name:"mia" }])
MongoDB sub query for JOIN--
`db.commentss.find().forEach(
function (newComments) {
newComments.userss = db.userss.find( { "uid": newComments.uid } ).toArray();
db.newCommentUsers.insert(newComments);
}
);`
Get result from newly generated Collection--
db.newCommentUsers.find().pretty()
Result--
`{
"_id" : ObjectId("5511236e29709afa03f226ef"),
"uid" : 12345,
"pid" : 444,
"comment" : "blah",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f2"),
"uid" : 12345,
"name" : "john"
}
]
}
{
"_id" : ObjectId("5511236e29709afa03f226f0"),
"uid" : 12345,
"pid" : 888,
"comment" : "asdf",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f2"),
"uid" : 12345,
"name" : "john"
}
]
}
{
"_id" : ObjectId("5511236e29709afa03f226f1"),
"uid" : 99999,
"pid" : 444,
"comment" : "qwer",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f3"),
"uid" : 99999,
"name" : "mia"
}
]
}`
Hope so this will help.
7
Why did you basically copy this nearly identical, one-year-old answer? stackoverflow.com/a/22739813/4186945
– hackel
May 5 '15 at 20:45
add a comment |
We can merge two collection by using mongoDB sub query. Here is example,
Commentss--
`db.commentss.insert([
{ uid:12345, pid:444, comment:"blah" },
{ uid:12345, pid:888, comment:"asdf" },
{ uid:99999, pid:444, comment:"qwer" }])`
Userss--
db.userss.insert([
{ uid:12345, name:"john" },
{ uid:99999, name:"mia" }])
MongoDB sub query for JOIN--
`db.commentss.find().forEach(
function (newComments) {
newComments.userss = db.userss.find( { "uid": newComments.uid } ).toArray();
db.newCommentUsers.insert(newComments);
}
);`
Get result from newly generated Collection--
db.newCommentUsers.find().pretty()
Result--
`{
"_id" : ObjectId("5511236e29709afa03f226ef"),
"uid" : 12345,
"pid" : 444,
"comment" : "blah",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f2"),
"uid" : 12345,
"name" : "john"
}
]
}
{
"_id" : ObjectId("5511236e29709afa03f226f0"),
"uid" : 12345,
"pid" : 888,
"comment" : "asdf",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f2"),
"uid" : 12345,
"name" : "john"
}
]
}
{
"_id" : ObjectId("5511236e29709afa03f226f1"),
"uid" : 99999,
"pid" : 444,
"comment" : "qwer",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f3"),
"uid" : 99999,
"name" : "mia"
}
]
}`
Hope so this will help.
We can merge two collection by using mongoDB sub query. Here is example,
Commentss--
`db.commentss.insert([
{ uid:12345, pid:444, comment:"blah" },
{ uid:12345, pid:888, comment:"asdf" },
{ uid:99999, pid:444, comment:"qwer" }])`
Userss--
db.userss.insert([
{ uid:12345, name:"john" },
{ uid:99999, name:"mia" }])
MongoDB sub query for JOIN--
`db.commentss.find().forEach(
function (newComments) {
newComments.userss = db.userss.find( { "uid": newComments.uid } ).toArray();
db.newCommentUsers.insert(newComments);
}
);`
Get result from newly generated Collection--
db.newCommentUsers.find().pretty()
Result--
`{
"_id" : ObjectId("5511236e29709afa03f226ef"),
"uid" : 12345,
"pid" : 444,
"comment" : "blah",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f2"),
"uid" : 12345,
"name" : "john"
}
]
}
{
"_id" : ObjectId("5511236e29709afa03f226f0"),
"uid" : 12345,
"pid" : 888,
"comment" : "asdf",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f2"),
"uid" : 12345,
"name" : "john"
}
]
}
{
"_id" : ObjectId("5511236e29709afa03f226f1"),
"uid" : 99999,
"pid" : 444,
"comment" : "qwer",
"userss" : [
{
"_id" : ObjectId("5511238129709afa03f226f3"),
"uid" : 99999,
"name" : "mia"
}
]
}`
Hope so this will help.
answered Mar 24 '15 at 9:27
KrishnaKrishna
11
11
7
Why did you basically copy this nearly identical, one-year-old answer? stackoverflow.com/a/22739813/4186945
– hackel
May 5 '15 at 20:45
add a comment |
7
Why did you basically copy this nearly identical, one-year-old answer? stackoverflow.com/a/22739813/4186945
– hackel
May 5 '15 at 20:45
7
7
Why did you basically copy this nearly identical, one-year-old answer? stackoverflow.com/a/22739813/4186945
– hackel
May 5 '15 at 20:45
Why did you basically copy this nearly identical, one-year-old answer? stackoverflow.com/a/22739813/4186945
– hackel
May 5 '15 at 20:45
add a comment |
protected by Community♦ Jun 29 '15 at 9:46
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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30
The last answer on this question is probably the most relevant, since MongoDB 3.2+ implemented a join solution called $lookup. Thought I would push it here because maybe not everyone will read to the bottom. stackoverflow.com/a/33511166/2593330
– thefourtheye
Nov 22 '15 at 16:33
6
Correct, $lookup was introduced in MongoDB 3.2. Details can be found at docs.mongodb.org/master/reference/operator/aggregation/lookup/…
– NDB
Nov 23 '15 at 13:20
i wonder why do not accept this anwser or any comment: stackoverflow.com/a/33511166/6941294
– Nozar Safari
Dec 18 '17 at 6:26