Hi Andrea,
> I am really new at PHP & MySQL, so please bear with me.
welcome to our happy band...
> I am using the mysql_insert_id function but continually receive an error,
> and hoping someone can point me in the right direction to resolve this.
>
> My error:
> Warning: Supplied argument is not a valid MySQL-Link resource on line 44
> _________________________________________
> $result = mysql_query("insert into support (ref_number, date_time, name,
> location, email, subject, problem, priority_1,
> priority_2, priority_3, priority_4) values ('null', '$date', '$name',
> '$email', '$location', '$subject', '$problem',
> '$priority_1', '$priority_2', '$priority_3', '$priority_4')");
>
> $ref_number = mysql_insert_id($link_id); ========> This is line 44
> ____________________________________________
> Any suggestions? The auto_num field in the database is updating and
> everything else is working fine....
Either remove $link_id and go with the default, or make sure that $link_id is the same
variable as assigned in
the mysql_connect() or mysql_pconnect() call - these being the functions that
establish/name the link.
Regards,
=dn
---------------------------------------------------------------------
Before posting, please check:
http://www.mysql.com/manual.php (the manual)
http://lists.mysql.com/ (the list archive)
To request this thread, e-mail <[EMAIL PROTECTED]>
To unsubscribe, e-mail <[EMAIL PROTECTED]>
Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
