代码无法正常工作，将MySQL结果排序为页面

Question

Ok so iv got this code which is suppose to bring back all my results from my database and display 5 results per page. At the moment I have 12 results in my database and it is only showing 10, it wont show the 3rd page as there isnt 5 results to display. Here is my code

//$_GET['page'] is to get the current page opened
if (isset($_GET["page"])) { $page  = $_GET["page"]; } else { $page=1; };
$results_per_page = "5"; //number of results I want displayed per page
$start_from = ($page-1) * $results_per_page; 
$query = "SELECT * FROM items WHERE subcat = '$conditions' LIMIT $start_from, $results_per_page";
$result = mysql_query($query) or die("Unable to verify user because : " . mysql_error());
while ($row_condition=mysql_fetch_array($result)) {
   //Display the results here
}

//Count number of results from database and work out how
//many pages I need to display all the results.    
$result1 = mysql_query("SELECT * FROM items WHERE subcat = '$conditions'");
$num_rows = mysql_num_rows($result1);
$num_pages = $num_rows / $results_per_page;

if ($num_rows > $results_per_page){
?>
<div id="pagenum">
<?php
//This creates and displays the page numbers for the user to select
foreach( range( 1, $num_pages) as $i) {
     if ($thepage == $i){
           echo '<b><a href="browse.php?condition=' . $condition . '&page=' . $i . '">' . $i . '</a></b>';
     }else{
           echo '<a href="browse.php?condition=' . $condition . '&page=' . $i . '">' . $i . '</a>';
     }

     //This places a line between each number of pages
     if ($i == $num_pages){
     }else{
         echo " | ";
     }
     }
?>
</div>
<?php
     }else{ echo "Test";}
?>

So can anyone see a problem with this?, so I have 12 entries in my database. However im only getting 2 pages, 5 on each page and im not getting the 3rd page as there are only 2 results left and it seems to want 5 to finish it. Thanks

Answer 1

You need to use ceil function:

$num_pages = ceil($num_rows / $results_per_page);