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'"; $result = mysql_query($query); $row = mysql_fetch_array($result); $users[$status] = $row ; return $users[$status] ; } $status = 0 ; $data = getusers($status); var_dump($data);
echo $data; ?>
_________________________________________________________________
Use MSN Messenger to send music and pics to your friends http://messenger.msn.co.uk
-- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php