Query samenstellen

Ik kwam dit script tegen en vroeg me af hoe ik de query kan samenstellen om dit te laten werken:

<?php

function page_count($page, $max_pp, $query, $page_counter)
{
$photos = mysql_result(mysql_query($query),0);
$pages = ceil($photos / $max_pp);
$first_page = ($page == 1) ? 0 : 1;
$last_page = ($page == $pages) ? 0 : $pages;
$prev = ($page == 1) ? 0 : $page - 1;
$next = ($page == $pages) ? 0 : $page + 1;

$discount = ($page > $page_counter) ? $page_counter : $page - 1;
$count = ($page < $pages - $page_counter) ? $page_counter : $pages - $page;
for ($i = $page - $discount; $i <= $page + $count; $i++)
{
$page_nr[] = $i;
}
return array($page, $prev, $next, $page_nr, $first_page, $last_page);
}

/* gebruik */
$get_rows = "SELECT COUNT(*) AS NUM FROM tabel WHERE iets = 'iets'";
$page = (!isset($_GET['page'])) ? 1 : $_GET['page'];
$pages = page_count($page, 10, $get_rows, 5);

$prev_lnk = ($pages[1] != 0) ? '<a href=?page='.$pages[1].'>'.$lang['link']['prev'].'</a>' : $lang['link']['prev'];
$next_lnk = ($pages[2] != 0) ? '<a href=?page='.$pages[2].'>'.$lang['link']['next'].'</a>' : $lang['link']['next'];
$first = ($pages[4] != 0) ? '<a href=?page='.$pages[4].'>'.$lang['link']['first'].'</a>' : $lang['link']['first'];
$last = ($pages[5] != 0) ? '<a href=?page='.$pages[5].'>'.$lang['link']['last'].'</a>' : $lang['link']['last'];

$page_numbers = '';
for ($i = 0; $i <= count($pages[3]) - 1; $i++)
{
$page_numbers .= ($pages[0] == $pages[3][$i]) ? '<b>['.$pages[3][$i].']</b>' : '<a href="?page='.$pages[3][$i].'">['.$pages[3][$i].']</a>';
}
?>

Het gaat me dan om dit stukje:
<?php
$get_rows = "SELECT COUNT(*) AS NUM FROM tabel WHERE iets = 'iets'";
?>