uh huh. o.o;
Sorry, that's one thing I'm not sure how to do. Here's my block of code:
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( $result );
Right now, it's giving errors saying:
Quote:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/prophot/public_html/uploading/viewgallery.php on line 27
Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in /home/prophot/public_html/uploading/viewgallery.php on line 31
But if I take off "WHERE customer = '".$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. <=/
02-08-2006, 08:15 PM
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