Insert data into a column if no data in column
0
How do I insert data into a column if that column has no values? This piece of code will not insert any values, unless I take out the if statement. Of course that isn't what I want, since that means each time the code is run, more data will be inserted into the column 'datagroup'.
include('dg_config.php');
# select column datagroup from table data
$stmt = $db -> prepare("SELECT datagroup FROM data");
$stmt->execute();
# fetch rows from data
while($row = $stmt->fetch()){
if (empty($row['datagroup'])){
$insertdg = "INSERT INTO data (datagroup)
VALUES ('Data Definitions'),('Policies'),('Systems and Process Documents'),('Purchase Orders'),('Invoices')";
$db ->exec($insertdg);
}
}
P.S dg_config.php contains the variable $db which is used for MySQL connection. Also, I have no values in the column so UPDATE won't work.
Desired output:
php mysql sql-insert
asked Nov 21 '18 at 14:08
wei123wei123
456
add a comment |
0
How do I insert data into a column if that column has no values? This piece of code will not insert any values, unless I take out the if statement. Of course that isn't what I want, since that means each time the code is run, more data will be inserted into the column 'datagroup'.
include('dg_config.php');
# select column datagroup from table data
$stmt = $db -> prepare("SELECT datagroup FROM data");
$stmt->execute();
# fetch rows from data
while($row = $stmt->fetch()){
if (empty($row['datagroup'])){
$insertdg = "INSERT INTO data (datagroup)
VALUES ('Data Definitions'),('Policies'),('Systems and Process Documents'),('Purchase Orders'),('Invoices')";
$db ->exec($insertdg);
}
}
P.S dg_config.php contains the variable $db which is used for MySQL connection. Also, I have no values in the column so UPDATE won't work.
Desired output:
php mysql sql-insert
asked Nov 21 '18 at 14:08
wei123wei123
456
did you try if ($row['datagroup']==''){
– Patrick Simard
Nov 21 '18 at 14:11
You have more values than columns.
– Funk Forty Niner
Nov 21 '18 at 14:13
@FunkFortyNiner think they are multiple rows.
– Nigel Ren
Nov 21 '18 at 14:14
@NigelRen I think they want to concat them into 1 column. Hard to say really what they want to do here. Question's unclear IMHO.
– Funk Forty Niner
Nov 21 '18 at 14:14
Not sure what you want. A table is made out of columns and then has multiple rows. How you mean: one column only once? Put a unique key on the column? Do you realise that you're inserting multiple rows into the table? And check your naming. Very confusing to have a table name and column with the same name.
– Jørgen
Nov 21 '18 at 14:14
add a comment |
0
0
0
How do I insert data into a column if that column has no values? This piece of code will not insert any values, unless I take out the if statement. Of course that isn't what I want, since that means each time the code is run, more data will be inserted into the column 'datagroup'.
include('dg_config.php');
# select column datagroup from table data
$stmt = $db -> prepare("SELECT datagroup FROM data");
$stmt->execute();
# fetch rows from data
while($row = $stmt->fetch()){
if (empty($row['datagroup'])){
$insertdg = "INSERT INTO data (datagroup)
VALUES ('Data Definitions'),('Policies'),('Systems and Process Documents'),('Purchase Orders'),('Invoices')";
$db ->exec($insertdg);
}
}
P.S dg_config.php contains the variable $db which is used for MySQL connection. Also, I have no values in the column so UPDATE won't work.
Desired output:
php mysql sql-insert
asked Nov 21 '18 at 14:08
wei123wei123
456
How do I insert data into a column if that column has no values? This piece of code will not insert any values, unless I take out the if statement. Of course that isn't what I want, since that means each time the code is run, more data will be inserted into the column 'datagroup'.
include('dg_config.php');
# select column datagroup from table data
$stmt = $db -> prepare("SELECT datagroup FROM data");
$stmt->execute();
# fetch rows from data
while($row = $stmt->fetch()){
if (empty($row['datagroup'])){
$insertdg = "INSERT INTO data (datagroup)
VALUES ('Data Definitions'),('Policies'),('Systems and Process Documents'),('Purchase Orders'),('Invoices')";
$db ->exec($insertdg);
}
}
P.S dg_config.php contains the variable $db which is used for MySQL connection. Also, I have no values in the column so UPDATE won't work.
Desired output:
php mysql sql-insert
php mysql sql-insert
asked Nov 21 '18 at 14:08
wei123wei123
456
asked Nov 21 '18 at 14:08
wei123wei123
456
edited Nov 21 '18 at 14:24
wei123
asked Nov 21 '18 at 14:08
wei123wei123
456
asked Nov 21 '18 at 14:08
wei123wei123
456
asked Nov 21 '18 at 14:08
wei123wei123
456
456
did you try if ($row['datagroup']==''){
– Patrick Simard
Nov 21 '18 at 14:11
You have more values than columns.
– Funk Forty Niner
Nov 21 '18 at 14:13
@FunkFortyNiner think they are multiple rows.
– Nigel Ren
Nov 21 '18 at 14:14
@NigelRen I think they want to concat them into 1 column. Hard to say really what they want to do here. Question's unclear IMHO.
– Funk Forty Niner
Nov 21 '18 at 14:14
Not sure what you want. A table is made out of columns and then has multiple rows. How you mean: one column only once? Put a unique key on the column? Do you realise that you're inserting multiple rows into the table? And check your naming. Very confusing to have a table name and column with the same name.
– Jørgen
Nov 21 '18 at 14:14
add a comment |
did you try if ($row['datagroup']==''){
– Patrick Simard
Nov 21 '18 at 14:11
You have more values than columns.
– Funk Forty Niner
Nov 21 '18 at 14:13
@FunkFortyNiner think they are multiple rows.
– Nigel Ren
Nov 21 '18 at 14:14
@NigelRen I think they want to concat them into 1 column. Hard to say really what they want to do here. Question's unclear IMHO.
– Funk Forty Niner
Nov 21 '18 at 14:14
Not sure what you want. A table is made out of columns and then has multiple rows. How you mean: one column only once? Put a unique key on the column? Do you realise that you're inserting multiple rows into the table? And check your naming. Very confusing to have a table name and column with the same name.
– Jørgen
Nov 21 '18 at 14:14
did you try if ($row['datagroup']==''){
– Patrick Simard
Nov 21 '18 at 14:11
did you try if ($row['datagroup']==''){
– Patrick Simard
Nov 21 '18 at 14:11
You have more values than columns.
– Funk Forty Niner
Nov 21 '18 at 14:13
You have more values than columns.
– Funk Forty Niner
Nov 21 '18 at 14:13
@FunkFortyNiner think they are multiple rows.
– Nigel Ren
Nov 21 '18 at 14:14
@FunkFortyNiner think they are multiple rows.
– Nigel Ren
Nov 21 '18 at 14:14
@NigelRen I think they want to concat them into 1 column. Hard to say really what they want to do here. Question's unclear IMHO.
– Funk Forty Niner
Nov 21 '18 at 14:14
@NigelRen I think they want to concat them into 1 column. Hard to say really what they want to do here. Question's unclear IMHO.
– Funk Forty Niner
Nov 21 '18 at 14:14
Not sure what you want. A table is made out of columns and then has multiple rows. How you mean: one column only once? Put a unique key on the column? Do you realise that you're inserting multiple rows into the table? And check your naming. Very confusing to have a table name and column with the same name.
– Jørgen
Nov 21 '18 at 14:14
Not sure what you want. A table is made out of columns and then has multiple rows. How you mean: one column only once? Put a unique key on the column? Do you realise that you're inserting multiple rows into the table? And check your naming. Very confusing to have a table name and column with the same name.
– Jørgen
Nov 21 '18 at 14:14
add a comment |
1 Answer
1
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oldest
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If you want to check if there are no records first and only insert the values then, a quick way from where you are is to say if the fetch() fails to retrieve a row then do the insert...
include('dg_config.php');
# select column datagroup from table data
$stmt = $db -> prepare("SELECT datagroup FROM data");
$stmt->execute();
if ( !$stmt->fetch() ){
$insertdg = "INSERT INTO data (datagroup)
VALUES ('Data Definitions'),('Policies'),('Systems and Process Documents'),('Purchase Orders'),('Invoices')";
$db ->exec($insertdg);
}
// re fetch the data
$stmt->execute();
This removes the while() loop altogether.
answered Nov 21 '18 at 14:27
Nigel RenNigel Ren
25.9k61832
the data does get inserted into the database but when I try to display it on the table on my web browser, the record 'Data Definitions' is missing. I inserted this codewhile($row = $stmt->fetch()){ ?> <tr> <td><?php echo $row['datagroup'] ?></td> <td><?php echo $row['dataowner']?></td> <td><?php echo $row['linkedpg']?></td> </tr> <?php } ?>after your code.
– wei123
Nov 21 '18 at 14:52
Can you check on the database if the value is there
– Nigel Ren
Nov 21 '18 at 14:53
the values are there but I can't retrieve the first row: link
– wei123
Nov 21 '18 at 14:55
I've altered the answer slightly, this should re start the fetch.
– Nigel Ren
Nov 21 '18 at 15:02
Ah thanks! I got it working now.
– wei123
Nov 21 '18 at 15:06
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
active
oldest
votes
0
If you want to check if there are no records first and only insert the values then, a quick way from where you are is to say if the fetch() fails to retrieve a row then do the insert...
include('dg_config.php');
# select column datagroup from table data
$stmt = $db -> prepare("SELECT datagroup FROM data");
$stmt->execute();
if ( !$stmt->fetch() ){
$insertdg = "INSERT INTO data (datagroup)
VALUES ('Data Definitions'),('Policies'),('Systems and Process Documents'),('Purchase Orders'),('Invoices')";
$db ->exec($insertdg);
}
// re fetch the data
$stmt->execute();
This removes the while() loop altogether.
answered Nov 21 '18 at 14:27
Nigel RenNigel Ren
25.9k61832
the data does get inserted into the database but when I try to display it on the table on my web browser, the record 'Data Definitions' is missing. I inserted this codewhile($row = $stmt->fetch()){ ?> <tr> <td><?php echo $row['datagroup'] ?></td> <td><?php echo $row['dataowner']?></td> <td><?php echo $row['linkedpg']?></td> </tr> <?php } ?>after your code.
– wei123
Nov 21 '18 at 14:52
Can you check on the database if the value is there
– Nigel Ren
Nov 21 '18 at 14:53
the values are there but I can't retrieve the first row: link
– wei123
Nov 21 '18 at 14:55
I've altered the answer slightly, this should re start the fetch.
– Nigel Ren
Nov 21 '18 at 15:02
Ah thanks! I got it working now.
– wei123
Nov 21 '18 at 15:06
add a comment |
0
If you want to check if there are no records first and only insert the values then, a quick way from where you are is to say if the fetch() fails to retrieve a row then do the insert...
include('dg_config.php');
# select column datagroup from table data
$stmt = $db -> prepare("SELECT datagroup FROM data");
$stmt->execute();
if ( !$stmt->fetch() ){
$insertdg = "INSERT INTO data (datagroup)
VALUES ('Data Definitions'),('Policies'),('Systems and Process Documents'),('Purchase Orders'),('Invoices')";
$db ->exec($insertdg);
}
// re fetch the data
$stmt->execute();
This removes the while() loop altogether.
answered Nov 21 '18 at 14:27
Nigel RenNigel Ren
25.9k61832
the data does get inserted into the database but when I try to display it on the table on my web browser, the record 'Data Definitions' is missing. I inserted this codewhile($row = $stmt->fetch()){ ?> <tr> <td><?php echo $row['datagroup'] ?></td> <td><?php echo $row['dataowner']?></td> <td><?php echo $row['linkedpg']?></td> </tr> <?php } ?>after your code.
– wei123
Nov 21 '18 at 14:52
Can you check on the database if the value is there
– Nigel Ren
Nov 21 '18 at 14:53
the values are there but I can't retrieve the first row: link
– wei123
Nov 21 '18 at 14:55
I've altered the answer slightly, this should re start the fetch.
– Nigel Ren
Nov 21 '18 at 15:02
Ah thanks! I got it working now.
– wei123
Nov 21 '18 at 15:06
add a comment |
0
0
0
If you want to check if there are no records first and only insert the values then, a quick way from where you are is to say if the fetch() fails to retrieve a row then do the insert...
include('dg_config.php');
# select column datagroup from table data
$stmt = $db -> prepare("SELECT datagroup FROM data");
$stmt->execute();
if ( !$stmt->fetch() ){
$insertdg = "INSERT INTO data (datagroup)
VALUES ('Data Definitions'),('Policies'),('Systems and Process Documents'),('Purchase Orders'),('Invoices')";
$db ->exec($insertdg);
}
// re fetch the data
$stmt->execute();
This removes the while() loop altogether.
answered Nov 21 '18 at 14:27
Nigel RenNigel Ren
25.9k61832
If you want to check if there are no records first and only insert the values then, a quick way from where you are is to say if the fetch() fails to retrieve a row then do the insert...
include('dg_config.php');
# select column datagroup from table data
$stmt = $db -> prepare("SELECT datagroup FROM data");
$stmt->execute();
if ( !$stmt->fetch() ){
$insertdg = "INSERT INTO data (datagroup)
VALUES ('Data Definitions'),('Policies'),('Systems and Process Documents'),('Purchase Orders'),('Invoices')";
$db ->exec($insertdg);
}
// re fetch the data
$stmt->execute();
This removes the while() loop altogether.
answered Nov 21 '18 at 14:27
Nigel RenNigel Ren
25.9k61832
edited Nov 21 '18 at 15:01
answered Nov 21 '18 at 14:27
Nigel RenNigel Ren
25.9k61832
answered Nov 21 '18 at 14:27
Nigel RenNigel Ren
25.9k61832
answered Nov 21 '18 at 14:27
Nigel RenNigel Ren
25.9k61832
25.9k61832
the data does get inserted into the database but when I try to display it on the table on my web browser, the record 'Data Definitions' is missing. I inserted this codewhile($row = $stmt->fetch()){ ?> <tr> <td><?php echo $row['datagroup'] ?></td> <td><?php echo $row['dataowner']?></td> <td><?php echo $row['linkedpg']?></td> </tr> <?php } ?>after your code.
– wei123
Nov 21 '18 at 14:52
Can you check on the database if the value is there
– Nigel Ren
Nov 21 '18 at 14:53
the values are there but I can't retrieve the first row: link
– wei123
Nov 21 '18 at 14:55
I've altered the answer slightly, this should re start the fetch.
– Nigel Ren
Nov 21 '18 at 15:02
Ah thanks! I got it working now.
– wei123
Nov 21 '18 at 15:06
add a comment |
the data does get inserted into the database but when I try to display it on the table on my web browser, the record 'Data Definitions' is missing. I inserted this codewhile($row = $stmt->fetch()){ ?> <tr> <td><?php echo $row['datagroup'] ?></td> <td><?php echo $row['dataowner']?></td> <td><?php echo $row['linkedpg']?></td> </tr> <?php } ?>after your code.
– wei123
Nov 21 '18 at 14:52
Can you check on the database if the value is there
– Nigel Ren
Nov 21 '18 at 14:53
the values are there but I can't retrieve the first row: link
– wei123
Nov 21 '18 at 14:55
I've altered the answer slightly, this should re start the fetch.
– Nigel Ren
Nov 21 '18 at 15:02
Ah thanks! I got it working now.
– wei123
Nov 21 '18 at 15:06
the data does get inserted into the database but when I try to display it on the table on my web browser, the record 'Data Definitions' is missing. I inserted this code
while($row = $stmt->fetch()){ ?> <tr> <td><?php echo $row['datagroup'] ?></td> <td><?php echo $row['dataowner']?></td> <td><?php echo $row['linkedpg']?></td> </tr> <?php } ?> after your code.– wei123
Nov 21 '18 at 14:52
the data does get inserted into the database but when I try to display it on the table on my web browser, the record 'Data Definitions' is missing. I inserted this code
while($row = $stmt->fetch()){ ?> <tr> <td><?php echo $row['datagroup'] ?></td> <td><?php echo $row['dataowner']?></td> <td><?php echo $row['linkedpg']?></td> </tr> <?php } ?> after your code.– wei123
Nov 21 '18 at 14:52
Can you check on the database if the value is there
– Nigel Ren
Nov 21 '18 at 14:53
Can you check on the database if the value is there
– Nigel Ren
Nov 21 '18 at 14:53
the values are there but I can't retrieve the first row: link
– wei123
Nov 21 '18 at 14:55
the values are there but I can't retrieve the first row: link
– wei123
Nov 21 '18 at 14:55
I've altered the answer slightly, this should re start the fetch.
– Nigel Ren
Nov 21 '18 at 15:02
I've altered the answer slightly, this should re start the fetch.
– Nigel Ren
Nov 21 '18 at 15:02
Ah thanks! I got it working now.
– wei123
Nov 21 '18 at 15:06
Ah thanks! I got it working now.
– wei123
Nov 21 '18 at 15:06
add a comment |
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did you try if ($row['datagroup']==''){
– Patrick Simard
Nov 21 '18 at 14:11
You have more values than columns.
– Funk Forty Niner
Nov 21 '18 at 14:13
@FunkFortyNiner think they are multiple rows.
– Nigel Ren
Nov 21 '18 at 14:14
@NigelRen I think they want to concat them into 1 column. Hard to say really what they want to do here. Question's unclear IMHO.
– Funk Forty Niner
Nov 21 '18 at 14:14
Not sure what you want. A table is made out of columns and then has multiple rows. How you mean: one column only once? Put a unique key on the column? Do you realise that you're inserting multiple rows into the table? And check your naming. Very confusing to have a table name and column with the same name.
– Jørgen
Nov 21 '18 at 14:14