> I am trying to write a few functions for a project i need to do for > school, > this function should return all the inactive, or active users (1 or 0 > staffstatusid) as an array for creating a drop down menu. > > Now when i perform the following code (yes it is included in a script that > connects to the databas i am using) it returns the following error. > > "Warning: Supplied argument is not a valid MySQL result resource in > /home/filterseveuk/public_html/project/getusers.php on line 7" > > and underneath the error it prints the value of var_dump($data) > which is " NULL" > > any ideas how to fix this? > > > ================================================================= > <? > function getusers($status){ > > $query = "SELECT FROM Staff WHERE Staffstatusid = '$status'"; Select _WHAT_ from the table? You do not have a valid query... > $result = mysql_query($query); > $row = mysql_fetch_array($result); > $users[$status] = $row ; > return $users[$status] ; > } > $status = 0 ; > $data = getusers($status); > var_dump($data); > > echo $data; > ?> ---John W. Holmes... PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php