Instead of checking if(!$id), perhaps you would be better off to check the
result of your query (which in this example was successful, since you got a
return from mysql() ).
Also, what is the last_insert_id() function that you are referencing in your
$insertIDQuery string?
As for things getting messed up when two queries hit the system at the same
time, you shouldn’t need to worry about that either because the two people
coming in would have different values for the $link variable and since
mysql_insert_id goes off of the link identifier of the connection (which
will be different for the two different people running queries), you should
not have a conflict.
HTH
Sam Masiello
Systems Analyst
Chek.Com
(716) 853-1362 x289
[EMAIL PROTECTED]
-----Original Message-----
From: Boget, Chris [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 24, 2001 4:49 PM
To: 'Sam Masiello'; Php (E-mail)
Subject: RE: [PHP] mysql_insert_id()
> You probably stated this in your previous post, but what is
> the result from your call to mysql() ? Is this call failing so
> that when you get to mysql_insert_id(), the id doesn't exist?
I did. It's returning a numerical one (1).
What I'm doing now is as follows. It's getting me the value
but I'm so afraid that if 2 queries get sent almost simultaneously
it's going to screw everything up...
$link = mysql_pconnect( "localhost", "thisuser", "thatpass" );
$result = mysql( $dbname, $dbquery );
$id = mysql_insert_id( $link );
if( !$id ) { $id = mysql_insert_id(); }
if( !$id )
$insertIDQuery = "SELECT last_insert_id()";
// get values from above query, etc., etc.
}
You get the gist of it... I cannot believe this stupid thing won't
work otherwise... And it's not consistent when it works and
doesn't. Very odd.
*sigh*
Chris
