RE: [PHP-DB] Join causing Error?

Where is the join condition on that quey? (the field in common on the to tables, to make the join). idsarts -----Mensaje original----- De: Keith Spiller [mailto:larentium@gilanet.com] Enviado el: Domingo, 25 de Febrero de 2001 05:30 a.m. Para: php-db@lists.php.net; php-general@lists.php.net CC: mysql@lists.mysql.com Asunto: [PHP-DB] Join causing Error? Can anyone tell me why this: Line 282 mysql_select_db("centraldb",$db); Line 283 $qorder++; Line 284 $result = mysql_query("SELECT q.questid, q.question, q.answer, q.qorder, q.depart, q.catid, q.active, q.global, q.adate, q.author, q.authoremail, q.askemail, c.catid, c.category, c.under, c.corder, c.active FROM central_groupfaqq q, central_groupfaqcat c WHERE q.active = '1' AND q.global = '1' AND c.active = '1' ORDER BY c.under, c.order, q.qorder",$db); Line 285 while ($myrow = mysql_fetch_row($result)) Would cause this error: Warning: Supplied argument is not a valid MySQL result resource in faqbody.php3 on line 285 When changing the same SELECT statement to: Line 284 $result = mysql_query("SELECT * FROM central_groupfaqq WHERE active = '1' ORDER BY qorder",$db); Works perfectly? Keith Spiller a.k.a. Larentium