Maybe a dumb question, but are you sure you have the fields id, songname and artist in a table named rage in a database named rage?
The times I got that error was because a non-existent field or a non-existent table. Otherwise, everything looks ok. Alex. ----- Original Message ----- From: "Mason Batley" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Thursday, January 31, 2002 6:48 AM Subject: [PHP] mysql G'day I keep getting an error message of :Warning: Supplied argument is not a valid MySQL result resource in /var/www/html/RageProject/StripPlayList.php on line 100 Rows: $link = mysql_connect("127.0.0.1", "rage", "rage"); mysql_select_db("rage"); $query = "Select id from songs where songname = '$Song' and artist = '$Artist'"; $result = mysql_query($query); $rows = mysql_num_rows($result); <<< Line 100 echo "Rows: $rows"; mysql_close($link); Can someone explain why.. I've been trying to work it out for 2 long. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]