Re: in_array()/finding page Problem

You solution is quite resource expensive. I would do: [find $maxpages] SELECT COUNT(*)/12 FROM table WHERE id <= $real_id; list($maxpages) = fetch_row() In result.php use "ORDER BY id" Ben G. McCullough wrote:

I think I have a flaw of logic in trying to find the correct page in a mutli-page sql result. Goal - find the correct page [results.php?page=x] when linking from another page. Current method - loop through a pagination function looking for the matching $id in an array [simplified for illustration]: [find $maxpages] $page =1; $catch=array(); while($page <= $maxpages) {

    [set start #]
    $result = [get sql results - limit  $startnumber, 12]
    while($record = mysql_fetch_array($result)) {
        extract($record);
        $catch[] = $found_id;
        }
    if(in_array($real_id, $catch, TRUE)) {
        $here = $page;
        }
    $page++;
    }

echo "results.php?page=$here"; I never seem to get a "true" return on in_array(), and in testing, I don't seem to be getting a full result set with my look up. I feel that I am missing something in the logic - or I am approaching this from the wrong direction. I originally wanted to do the while test as "if in_array is FALSE, continue loop" but I couldn't get that to work at all. Thank you for any help you can give this newbie.