Aggregate query result in mongodb
0
I have collection with documents like this one:
{
"_id": 1,
"people": [
{
"name": "Bob",
"age": "15"
},
{
"name": "Alice",
"age": "18"
}
]
}
My query is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "people.name": 1 } })
This query returns:
{
"people": [
{
"name": "Bob"
},
{
"name": "Alice"
}
]
}
But I need the result like:
{ "names": [ "Bob", "Alice" ] }
Which parameters should I add to the .aggregate() function?
mongodb aggregate
asked Nov 23 '18 at 14:39
TheYarikTheYarik
156
add a comment |
0
I have collection with documents like this one:
{
"_id": 1,
"people": [
{
"name": "Bob",
"age": "15"
},
{
"name": "Alice",
"age": "18"
}
]
}
My query is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "people.name": 1 } })
This query returns:
{
"people": [
{
"name": "Bob"
},
{
"name": "Alice"
}
]
}
But I need the result like:
{ "names": [ "Bob", "Alice" ] }
Which parameters should I add to the .aggregate() function?
mongodb aggregate
asked Nov 23 '18 at 14:39
TheYarikTheYarik
156
1
No need to use$maphere... Try this.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
– Anthony Winzlet
Nov 23 '18 at 15:16
Thanks! It's helped
– TheYarik
Nov 23 '18 at 15:22
add a comment |
0
0
0
0
I have collection with documents like this one:
{
"_id": 1,
"people": [
{
"name": "Bob",
"age": "15"
},
{
"name": "Alice",
"age": "18"
}
]
}
My query is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "people.name": 1 } })
This query returns:
{
"people": [
{
"name": "Bob"
},
{
"name": "Alice"
}
]
}
But I need the result like:
{ "names": [ "Bob", "Alice" ] }
Which parameters should I add to the .aggregate() function?
mongodb aggregate
asked Nov 23 '18 at 14:39
TheYarikTheYarik
156
I have collection with documents like this one:
{
"_id": 1,
"people": [
{
"name": "Bob",
"age": "15"
},
{
"name": "Alice",
"age": "18"
}
]
}
My query is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "people.name": 1 } })
This query returns:
{
"people": [
{
"name": "Bob"
},
{
"name": "Alice"
}
]
}
But I need the result like:
{ "names": [ "Bob", "Alice" ] }
Which parameters should I add to the .aggregate() function?
mongodb aggregate
mongodb aggregate
asked Nov 23 '18 at 14:39
TheYarikTheYarik
156
asked Nov 23 '18 at 14:39
TheYarikTheYarik
156
asked Nov 23 '18 at 14:39
TheYarikTheYarik
156
asked Nov 23 '18 at 14:39
TheYarikTheYarik
156
asked Nov 23 '18 at 14:39
TheYarikTheYarik
156
156
1
No need to use$maphere... Try this.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
– Anthony Winzlet
Nov 23 '18 at 15:16
Thanks! It's helped
– TheYarik
Nov 23 '18 at 15:22
add a comment |
1
No need to use$maphere... Try this.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
– Anthony Winzlet
Nov 23 '18 at 15:16
Thanks! It's helped
– TheYarik
Nov 23 '18 at 15:22
1
1
No need to use
$map here... Try this .groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })– Anthony Winzlet
Nov 23 '18 at 15:16
No need to use
$map here... Try this .groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })– Anthony Winzlet
Nov 23 '18 at 15:16
Thanks! It's helped
– TheYarik
Nov 23 '18 at 15:22
Thanks! It's helped
– TheYarik
Nov 23 '18 at 15:22
add a comment |
1 Answer
1
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oldest
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0
The solution is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
answered Nov 23 '18 at 15:17
TheYarikTheYarik
156
add a comment |
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0
The solution is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
answered Nov 23 '18 at 15:17
TheYarikTheYarik
156
add a comment |
0
The solution is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
answered Nov 23 '18 at 15:17
TheYarikTheYarik
156
add a comment |
0
0
0
The solution is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
answered Nov 23 '18 at 15:17
TheYarikTheYarik
156
The solution is:
db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
answered Nov 23 '18 at 15:17
TheYarikTheYarik
156
edited Nov 23 '18 at 15:23
answered Nov 23 '18 at 15:17
TheYarikTheYarik
156
answered Nov 23 '18 at 15:17
TheYarikTheYarik
156
answered Nov 23 '18 at 15:17
TheYarikTheYarik
156
156
add a comment |
add a comment |
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1
No need to use
$maphere... Try this.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })– Anthony Winzlet
Nov 23 '18 at 15:16
Thanks! It's helped
– TheYarik
Nov 23 '18 at 15:22