On Tuesday 30 October 2001 15:49, TorrentUK wrote: > Please could some take a look at this code and tell me why > when I take my IF statements out of the function and put in > them in the same place where I call the function from they > work, but as soon as I replace them with the function name > they don't? > > Appreciate any help. > torrent > > Here's the code...
Your variable is not in scope. Either pass your variable as a reference in the function definition: function RatingFilter (&$sql) { ... } or use the global keyword to bring your variable into scope: function RatingFilter () { global $sql; if ($br) {$sql.= " and beg_rate >= '2'";} ... } And follow the advice of Charles and read up on it so you know how they are different. -- PHP Database 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]