我正在努力使跳转形式程序化为面向对象的样式，因此如果我的代码不整洁或有缺陷，请注意-在这里，我通过jquery将一些帖子传递给一个类，以便在用户选中复选框时更新记录：
这是数据库连接

class db {
    private $host ;      
    private $username;
    private $password;
    private $dbname;
    protected function conn()
    {
        $this->host = "localhost";
        $this->username = "root";
        $this->password = "";
        $this->dbname = "mytest";
        $db = new mysqli($this->host, $this->username,  $this->password, $this->dbname);
        if($db->connect_errno > 0){
                die('Unable to connect to database [' . $db->connect_error . ']');
        }
        return $db;
    }
}

这是更新类

class updOrders extends db {
public $pid;
public $proc;
public function __construct()
{
$this->pid = isset($_POST['pid']) ? $_POST['pid'] : 0;
$this->proc = isset($_POST['proc']) ? $_POST['proc'] : 1;
// $stmt = $this->conn()->query("UPDATE tblorderhdr SET completed = ".$this->proc." WHERE orderid = ".$this->pid);
$stmt = $this->conn()->prepare("UPDATE tblorderhdr SET completed = ? WHERE orderid = ?");
$stmt->bind_param('ii', $this->proc, $this->pid);
 $stmt->execute();
if($stmt->error)
    {
        $err = $stmt->error ;
    } else {
        $err = 'ok';
    }
/* close statement */
$stmt->close();
 echo json_encode($err);   
}
}
$test = new  updOrders;

当我注解掉prepare语句并直接运行查询（注解掉）时，它会更新，当我尝试将其作为prepare语句运行时，它会返回一个错误“mysql server has gone away”。
任何帮助都非常感谢我已经没有主意了！
ps:localhost上的apache/xampp