mysqli prepare语句在执行时只返回部分结果，即在fetch（）$hash，$type保留其值，但$userid保持为空，此问题只出现在prepared语句中。

if($statement=$this->db->prepare("SELECT `id`,`password`,`type` FROM `login` WHERE `username`=?")){
   $statement->bind_param("s",$username);
   if($statement->execute()){
     $statement->bind_result($userId,$hash,$type);
     if($statement->fetch()){
       if(password_verify($password, $hash)){
         $this->userId=$userId;
         if($type=='u')
            return true;
       }else {
         $this->error(11);
       }
     }else
       $this->error(5);
   }else
    $this->error(4);
 }else
   $this->error(3);