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;
?>
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