PHP引用函数的奇怪行为

我有一个引用脚本这个脚本

<?php
ob_start();
define('DB_HOST', 'localhost');
define('DB_NAME', 'dbnamehere');
define('DB_USER', 'dbuserhere');
define('DB_PASS', 'dbpasshere');
mysql_connect(DB_HOST,DB_USER,DB_PASS);
mysql_select_db(DB_NAME);
$id = $_REQUEST['id']; 
$uid = $_REQUEST['uid']; 
$oid = $_REQUEST['oid']; // completed offer or payment method
$new = $_REQUEST['new']; 
$total = $_REQUEST['total'];
$sig = $_REQUEST['sig'];
$timestamp = date("Y-m-d H:i:s");
// Secrete Key
$key = 'e5870b6ab402d790a5d6bd1cefaee7c4';
// Compare results
$hash = md5($id.':'.$new.':'.$uid.':'.$key);
// Output results
if ($sig == $hash) {
        print "1'n";

//Users point update query here
 $users = mysql_query("SELECT points FROM users WHERE id=".$uid);
 $rows = mysql_fetch_array($users);
 $user_points = $rows['points'];
 $query1 = mysql_query("update users set points=($user_points+$new/2) where id=$uid ");

//Updating referral coins 
        $query2 = "select points, referral_id from users where referral_id=".$uid;  
        $user_rows = mysql_query($query2);
        $all=mysql_fetch_array($user_rows,MYSQL_BOTH);
        if($all['referral_id'] != 0){
             echo $referal_points = intval((25/100) * $new);
             $update_referral_points = "update users set points = points + '$referal_points' WHERE id = ".$all['referral_id'];
             mysql_query($update_referral_points);
                    }
        } else {
            print "0'n";
}
?>

当我运行这个脚本时，数据库行不更新，参见下面的示例

    id | points | referral_id
    ---|--------|--------
    1  | 1000   | 2
    2  | 2000   | 0
    3  | 1000   | 2

:

if $uid = 1 &$new = 100或$uid = 3 &$new = 100
我们需要将$new=(100*25)/100 = +25奖励给id=2中的id=2，因为id= 1 &2 .

    id | points | referral_id
    ---|--------|--------
    1  | 1100   | 2
    2  | 2000   | 0
    3  | 1100   | 2

成功后，我期待这样的结果

    id | points | referral_id
    ---|--------|--------
    1  | 1100   | 2
    2  | 2025   | 0
    3  | 1100   | 2

脚本成功打印"1'n";query1也工作没有query2，当我运行query2它停止更新数据库，甚至停止打印"1'n";