warning problem: expects parameter 1 to be mysqli_result [duplicate]

Answer 1

Waterfall is probably right. Revise your code as follows:

$result = mysqli_query($dbc,$sql1) or die(mysqli_error($dbc));
// and
$result = mysqli_query($dbc,$sql2) or die(mysqli_error($dbc));

PS: Just wondering what exactly is $page? Did you forget to do a:

global $page;