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);

echo $data;

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