Grabbing Single Value From Other Table To Insert

[Kristoff1875]

Hi guys, i'm currently using:

 

mysql_query("INSERT INTO Jobs (Username, JobNumber, Description, Status) VALUES ('$_POST[ClientUsername]', '$_POST[JobNumber]', '$_POST[Description]', '$_POST[status]')")
or die(mysql_error());

 

Which is inserting all the right values in to my database, only thing is I also want "ClientID" from Clients table to be fetched and put in like this:

 

mysql_query("INSERT INTO Jobs (Username, ClientID JobNumber, Description, Status) VALUES ('$_POST[ClientUsername]', $ClientID, '$_POST[JobNumber]', '$_POST[Description]', '$_POST[status]')")
or die(mysql_error());

 

As I say, this is on a different table. I've tried lots of different ways, like INSERT SELECT etc, but can't seem to have any joy.

 

Any help much appreciated.

 

Thanks

Edited December 18, 2012 by Kristoff1875

[Love2c0de]

What is the name of your other table?

 

I think you may have to do a join.

 

Regards,

 

AoTB.

[Kristoff1875]

The other table is called 'Clients'.

 

Cheers

Edited December 19, 2012 by Kristoff1875

[berridgeab]

You need an Insert Select syntax.

 

You only need the clientId from the other table, and Select will return strings so long as you have properly escaped them.

 

You will need to do some of the POST keys.

$query = "
INSERT INTO
Jobs (Username, ClientID, JobNumber, Description, Status)
SELECT
'{$_POST['someVar']}',
clientId,
'{$_POST['someVar']}',
'{$_POST['someVar']}',
'{$_POST['someVar']}'
FROM Clients
WHERE clientId = 2";

Edited December 19, 2012 by berridgeab

[Kristoff1875]

Should this be right?

 

INSERT INTO
Jobs (ClientID, Username, JobNumber, Description, Status)
SELECT
ClientID,
'{$_POST['ClientUsername']}',
'{$_POST['JobNumber']}',
'{$_POST['Description']}'
'{$_POST['Status']}'
FROM Clients
WHERE Username = '{$_POST['ClientUsername']}'"

 

Because I want it to find the ClientID based on the username that is posted. At the moment i'm getting:

 

Column count doesn't match value count at row 1

 

Ah hold on, missing comma!

 

With the comma it works. Thanks so much

Edited December 19, 2012 by Kristoff1875

[Muddy_Funster]

you really shouldn't be sending raw $_POST[] data anywhere neer your database