Thanks a lot. It was an error... a HUGE one though :-) César Aracena IS / MCSE+I Neuquén, NQN (0299) 156-356688 (0299) 446-6621
> -----Mensaje original----- > De: Denis Arh [mailto:[EMAIL PROTECTED]] > Enviado el: Domingo, 16 de Junio de 2002 04:28 a.m. > Para: [EMAIL PROTECTED] > Asunto: Re: [PHP-DB] Arraying JOINED tables > > Try find out if you have an error in your query, like this: > > $query = "SELECT * FROM logins,auth WHERE logins.authname = $variable1 > AND auth.authid = $variable2"; > if ( $result = @mysql_query($query) ) { > $row = mysql_fetch_array($result); > } > else { > echo mysql_error(); > } > > and the other thing... maybe you have fields in both table logins and auth > with the same name > > > Denis Arh > > > > ----- Original Message ----- > From: "César Aracena" <[EMAIL PROTECTED]> > To: "PHP DB List" <[EMAIL PROTECTED]> > Sent: Sunday, June 16, 2002 8:59 AM > Subject: [PHP-DB] Arraying JOINED tables > > > Hi all. Hope you're all alright since I don't see any of you writing for > some time now ;-) > > This should be an easy one for all of you. I want to make a basic SELECT > query from two tables and fetch all the results into one array. I'm > doing it like this: > > $query = "SELECT * FROM logins,auth WHERE logins.authname = $variable1 > AND auth.authid = $variable2"; > $result = mysql_query($query); > $row = mysql_fetch_array($result); > > and I get: Warning: Supplied argument is not a valid MySQL result > resource in blah blah. > > What am I doing wrong? Thanks in advance, > > Cesar Aracena <mailto:[EMAIL PROTECTED]> > CE / MCSE+I > Neuquen, Argentina > +54.299.6356688 > +54.299.4466621 > > > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
