Try:
$query = "SELECT songname FROM mp3 WHERE sgid = '" .$id."'";
an SQL query requires you to quote strings.
james
-----Original Message-----
From: Chris Cocuzzo [mailto:[EMAIL PROTECTED]]
Sent: Monday, July 23, 2001 6:57 PM
To: [EMAIL PROTECTED]
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
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