validasi banyak inputan

----------------- index.php ----------------
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Papermashup.com | jQuery PHP Username Checker</title>

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.js"></script>
<script>
$(document).ready(function(){
$('.username').keyup(username_check);
});

function username_check(){
var username = $('.username').val();
if(username == "" || username.length < 4){
$('.username').css('border', '3px #CCC solid');
$('.tick').hide();
}else{

jQuery.ajax({
   type: "POST",
   url: "check.php",
   data: 'username='+ username,
   cache: false,
   success: function(response){
if(response == 1){
    $('.username').css('border', '3px #C33 solid');
    $('.tick').hide();
    $('.cross').fadeIn();
    }else{
    $('.username').css('border', '3px #090 solid');
    $('.cross').hide();
    $('.tick').fadeIn();
         }

}
});
}
}
</script>

<style>
.username{
    padding:3px;
    font-size:18px;
    border:3px #CCC solid;
}

.tick{display:none}
.cross{display:none}


</style>
</head>

<body>

Username: <input name="username[]" rel='1' class="username" id="username" type="text" />
<img class="tick" src="check.gif" width="16" height="16"/>
<img class="cross" src="cross.png" width="16" height="16"/>
<br><br>
Username: <input name="username[]" rel='2' class="username" type="text" />
<img class="tick" src="check.gif" width="16" height="16"/>
<img class="cross" src="cross.png" width="16" height="16"/>

</body>
</html>

--------------------- check.php----------------
<?php
 // Get the main settings from the array we just loaded
        $host = 'localhost';
        $db = 'karyawan';
        $user = 'root';
        $pass = '';

        // Connect to the database
        mysql_connect($host, $user, $pass)or die('koneksi error ');
        mysql_select_db($db) or die('database error');

$username = trim(strtolower($_POST['username']));
$username = mysql_escape_string($username);

$result =mysql_query("SELECT nama FROM user WHERE nama = '$username' LIMIT 1");
$num = mysql_num_rows($result);
echo $num;
?>