ошибка в коде php

<?
 
include ("config.php");
$num=1; 
$page = (int)($_GET['page']);
$count = mysql_query("SELECT COUNT(*) FROM java WHERE visible ='1'", $link);
$temp = mysql_fetch_array($count);
if ($temp[0]>0)
{
    $tempcount=$temp[0];
    $total=(($tempcount -1)/$num)+1;
    $total=intval($total);
    $page=intval($page);
    if(empty($page) or $page<0) $page=1;
    if($page > $total) $page= $total;
    $start= $page * $num - $num;
    $query_start_num ="LIMIT $start, $num";
    $result = mysql_query("SELECT * FROM java WHERE visible='1' ORDER BY $query_start_num", $link);
    if(mysql_num_rows($result)>0){
    $row =mysql_fetch_array($result);
    do{
        echo '<div class="header_post"><div class="header_footer_left_post"><a href="java-'.$row["name_url"].'.html">'.$row["name"].'</a></div>
        <div class="header_footer_right_post"></div></div>';
        echo'<div class="footer_post">
        <div class="header_footer_post"><h3>';
        $symb =450;                  
        $text = $row['description'] ;                   
        $announce = substr($text, 0, $symb);
        echo $announce.'</h3></div>
        <div class="footer_footer_post"><a href="java-'.$row["name_url"].'.html"><img src="img/dalee.png"/></a></div>
        </div>';
        }
    
    while($row =mysql_fetch_array($result));
    }
}

if(mysql_num_rows($result)>0){

if (!$result) die("Я не умею писать запросы, сервер вернул такую ошибку: <p>". mysql_error() . "</p></br>мой запрос:<p>SELECT * FROM java WHERE visible='1' ORDER BY $query_start_num</p>");

$result = mysql_query("SELECT * FROM java WHERE visible='1' ORDER BY'".$query_start_num."'", $link);