Re: MySQL error, what's wrong here..

try this: $query = "SELECT songname FROM mp3 WHERE sgid = \"$id\""; (or a little cleaner: $query = sprintf("SELECT songname FROM mp3 WHERE sgid = \"%d\"", $id); ) instead of: $query = "SELECT songname FROM mp3 WHERE sgid = " .$id; ----- Original Message ----- From: "James Holloway" <james@towntalk.co.uk> To: <php-general@lists.php.net> Sent: Tuesday, July 24, 2001 11:05 AM Subject: Re: [PHP]MySQL error, what's wrong here.. > Hi Chris, > > If you're using MySQL 3.23+, you might want to consider using something > like: > > SELECT songname FROM mp3 ORDER BY RAND() LIMIT 1 > > Not that this answers your original problem, but it seems to make more sense > than manually coding a random number (which is, perhaps, impractical > especiallyif you plan to add / take away entries to your table on a regular > basis). > > James > > "Chris Cocuzzo" <cuzo@mediaone.net> wrote in message > news:014d01c113ca$dd3bf460$0300a8c0@mediaone.net... > > <?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 > > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net > For additional commands, e-mail: php-general-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net >