我有一个查询，它在phpmyadmin中生成结果，但在codeigniter中不生成结果。

$sql = "SELECT express_interests.*, 
       cl_to  .User_Name AS ToClient, 
       cl_from.User_Name AS FromClient,
       cl_from.Member_Id AS FromMid,
       cl_to.Member_Id AS ToMid

FROM express_interests  
    INNER JOIN users AS cl_to ON cl_to.User_Id = express_interests.To_Id
    INNER JOIN users AS cl_from ON cl_from.User_Id = express_interests.User_Id";

我想在codeignator中使用相同的查询。这就是我用过的

$this->db->select('express_interests.*, 
       cl_to  .User_Name AS ToClient, 
       cl_from.User_Name AS FromClient,
       cl_from.Member_Id AS FromMid,
       cl_to.Member_Id AS ToMid
');

$this->db->from('express_interests');

$this->db->join('users AS cl_to', 'cl_to.User_Id = express_interests.To_Id');
$this->db->join('users AS cl_from', 'cl_from.User_Id = express_interests.User_Id');

当我用这个的时候它说
“字段列表”中的未知列“cl\u to.user\u name”
在codeigniter中使用上述查询的正确方法是什么。