Re: New years resolution: To get serious with my programming! Anyone wanna help? :)

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> From: Jason Pruim <japruim@xxxxxxxxxx>
> Date: Tue, 8 Jan 2008 13:18:40 -0500
> To: Jack Mays <jmays@xxxxxxxxxxx>
> Cc: php-general General List <php-general@xxxxxxxxxxxxx>
> Subject: Re:  New years resolution: To get serious with my programming!
> Anyone wanna help? :)
> 
> 
> 
> Actually, the problem isn't the query... the query works fine. What I
> want is for if it doesn't work, I want it to tell the user that it
> didn't work.
> 
> Right now I'm just blindly accepting that the insert succeeded, and I
> want to get away from that! :)
> 
> Does that explain it a little better?

If you are simply wanting to do a check on whether the insert was successful
or not, then grab the id form the insert and check to see if it is there...

$success = mysql_insert_id();

If(!$success){ 
   echo "no joy.. Insert failed.";
} 

Because mysql_insert_id() acts on the last performed query, be sure to call
mysql_insert_id() immediately after the query that generates the value.


--
Stephen Johnson c | eh
The Lone Coder

http://www.thelonecoder.com
continuing the struggle against bad code

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