Unknown column 'userc.u_id' in 'on clause'

本文关键字：in on clause id userc column Unknown | 更新日期: 2023-09-27

我正在使用Codeigniter，我试图检索角色ID为3的所有用户及其所有保存的自定义字段。由于某种原因，我得到一个数据库错误

"未知列'userc。u_id' in 'on子句'"

   $this->db->join('(SELECT GROUP_CONCAT(data) AS custom_data, id AS dataid, u_id
     FROM ea_user_cfields userc
     GROUP BY id) AS tt', 'userc.u_id = ea.id','left');
    $this->db->join('(SELECT GROUP_CONCAT(name) AS custom_name, id AS customid
     FROM ea_customfields AS cf
     GROUP BY id) AS te', 'userc.c_id = cf.id','left');
    $this->db->where('id_roles', $customers_role_id);

    return $this->db->get('ea_users ea')->result_array();

实际查询:

SELECT * FROM (`ea_users` ea) 
LEFT JOIN (SELECT GROUP_CONCAT(data) AS custom_data, id AS dataid, u_id FROM
ea_user_cfields userc GROUP BY id) AS tt ON `userc`.`u_id` =
`ea`.`id` 
LEFT JOIN (SELECT GROUP_CONCAT(name) AS
custom_name, id AS customid FROM ea_customfields AS cf GROUP BY id) AS
te ON `userc`.`c_id` = `cf`.`id` WHERE `id_roles` = '3'

使用下面的代码可能会对您有所帮助。

SELECT * FROM (`ea_users` ea) 
LEFT JOIN (SELECT GROUP_CONCAT(data) AS custom_data, id AS dataid, u_id FROM
ea_user_cfields userc GROUP BY id) AS tt ON tt.`u_id` =
`ea`.`id` 
LEFT JOIN (SELECT GROUP_CONCAT(name) AS
custom_name, id AS customid FROM ea_customfields AS cf GROUP BY id) AS
te ON tt.`c_id` = `cf`.`id` WHERE `id_roles` = '3'