Hi,
I wrote code something like
$query=".............";
$result=@mysql_query($query) or die($php_errormsg);

I got a message to say
Notice: Undefined variable: php_errormsg in ...

Anybody can let me know what's wrong with it and how to fix it?
i changed track_errors=On in my php.ini

Thank you all,
Jun

You probably want this:
$result=@mysql_query($query) or die(mysql_error());

Regards,
Rich

--
[http://www.smilingsouls.net]
Mail_IMAP: A PHP/C-Client/PEAR solution for webmail
Author: Beginning CSS: Cascading Style Sheets For Web Design

i have used $result=@mysql_query($query) or die(mysql_error());
before, but some doc say that $php_errormsg can report more info in details.
http://php.mirrors.ilisys.com.au/man...rorcontrol.php have some info about $php_errormsg

Any idea?

Thanks,
Jun

That may be the case if you are calling on a PHP function and an error is generated, but you aren't going to see the database errors generated by MySQL using that method, which are more useful in this case. I didn't even know such a method existed, so needless to say I've never used it.

To get around the Notice error, add this to the start of your code. I assume that variable is only created if there is an error to report. For that bit of code you have there, mysql_error() is what you want to use anyway.

if (!isset($php_errormsg))
{
    $php_errormsg = '';
}

HTH!

Regards,
Rich

--
[http://www.smilingsouls.net]
Mail_IMAP: A PHP/C-Client/PEAR solution for webmail
Author: Beginning CSS: Cascading Style Sheets For Web Design

Yes, you are right. Forget about the $php_errormsg for now.

At the same time, do you have any way to see more info in details about mysql error?

Thanks,
Jun