RE: [PHP]MySQL error, what's wrong here..

my guess would be that sgid is a character field. if this isn't the case then try hardcoding a known id into the query and seeing what happens. It could also be that the connection has failed, try something like: if ($connection = mysql_connect()) { ... $id = rand(1,2); $query = "SELECT songname FROM mp3 WHERE sgid = " .$id; if ($result = mysql_query($query,$connection)) { $mp3d = mysql_fetch_array($result); ... } else echo("query failed"); ... } else echo("connection failed"); this'll give you an idea of where it's going wrong Tim Ward Senior Systems Engineer Please refer to the following disclaimer in respect of this message: http://www.stivesdirect.com/e-mail-disclaimer.html > -----Original Message----- > From: Chris Cocuzzo [mailto:cuzo@mediaone.net] > Sent: 23 July 2001 23:57 > To: php-general@lists.php.net > Subject: [PHP]MySQL error, what's wrong here.. > > > <?php > $id = rand(1,2); > $query = "SELECT songname FROM mp3 WHERE sgid = " .$id; > $result = mysql_query($query,$connection); > $mp3d = mysql_fetch_array($result); > ?> > > the server is telling me line 43(which starts with $mp3d) is > not a valid > mysql result resource. what am i doing wrong?? > > chris > >