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
Otherwise, everything looks ok.
----- Original Message -----
From: "Mason Batley" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, January 31, 2002 6:48 AM
Subject: [PHP] mysql
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
$link = mysql_connect("127.0.0.1", "rage", "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";
Can someone explain why.. I've been trying to work it out for 2 long.
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