[Red Blaze]

uh huh. o.o;
Sorry, that's one thing I'm not sure how to do. Here's my block of code:

		$result = mysql_query( "SELECT c.album_id,c.album_name,COUNT(photo_id)
						FROM gallery_albums
						WHERE customer = '".$customer."' as c
						LEFT JOIN gallery_photos as p ON p.photo_album = c.album_id
						GROUP BY c.album_id" );
line 27->		while( $row = mysql_fetch_array( $result ) )
		{
			$result_array[] = "<a href='viewgallery.php?aid=".$row[0]."'>".$row[1]."</a> "."(".$row[2].")";
		}
line 31->		mysql_free_result( $result );

Right now, it's giving errors saying:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/prophot/public_html/uploading/viewgallery.php on line 27

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in /home/prophot/public_html/uploading/viewgallery.php on line 31

But if I take off "WHERE customer = '".$customer."'", the errors go *poof*. I want to filter the albums by customer so only THAT logged in customer views his/her albums and not anyone elses. <=/