无法获得正确的值并打印相应的消息

Question

以下是我的代码，我正在尝试附加$vid的值，而不是下次我的文件加载时，我正在尝试查找数组$vid中当前传递的$ids值，但每次我获得成功的时候，我出错了，我不能正确追加价值吗？

process.php

session_start();
if(isset($_POST)) {
    $reason_data = trim($_POST['abuse_data']);
    if($reason_data ==  "Video is offensive")
    {
        $reason = 'a:1:{i:0;s:18:"' . $reason_data .'";}';
    }
    else 
    {
            $reason = 'a:1:{i:0;s:19:"' . $reason_data .'";}';
    }       
   $motive = trim($_POST['remark_data']);
   $uid = trim($_POST['abuse_uid']);
   $vid = trim($_POST['abuse_vid']);
   if(trim($_SESSION['abuse_vdo_id']) == '')
   { 
    $_SESSION['abuse_vdo_id'] = $vid;  
        $str = $_SESSION['abuse_vdo_id'];           
   }
   else {
    $str = $str."," .$vid;
    $_SESSION['abuse_vdo_id'] = $str;
   }
    print_r($_SESSION['abuse_vdo_id']);
    echo "break";
    print_r($str);
    $ids = explode(',',$_SESSION['abuse_vdo_id']);
    echo "break 2";
    print_r($ids);
   if(in_array($vid, $ids))
        {       
            echo "already abused";
        }  
        else
        {
            echo "success"; 
            $query = "INSERT INTO ".DB_PREFIX."reports ( reason, motive, uid, vid ) VALUES ( '{$reason}','{$motive}', '{$uid}', '{$vid}' )";                
            $result = mysql_query($query);   
        }
}