sql（mysql）中与旧用户相关的用户列表

我在数据库中有这个表：

user
-id
-name
-role_id (related to a role-table)

示例数据：

name of user1. id=1, role_id = 2, old_userid = 0
name of user2. id=2, role_id = 2, old_userid = 0
name of user3. id=3, role_id = 2, old_userid = 0
name of user4, id=4, role_id = 3, old_userid = 2
name of user5, id=5, role_id = 3, old_userid = 2
name of user6, id=6, role_id = 3, old_userid = 4

name of user2[user4, user5] //user4 and user5 has older_userid as 2
name of user6[user4]        //user6 has older_userid as 4

通过执行以下操作，我可以获得此旧用户列表：

$this->db->select('u.name')->from('user u');      
$this->db->where('u.old_userid >', '0');

它会回来的

name of user4  
name of user5
name of user6

但是我想要：

name of user1. id=1, role_id = 2, old_userid = 0
name of user2. id=2, role_id = 2, old_userid = 0
name of user3. id=3, role_id = 2, old_userid = 0
name of user4, id=4, role_id = 3, old_userid = 2
name of user5, id=5, role_id = 3, old_userid = 2
name of user6, id=6, role_id = 3, old_userid = 4

name of user2[user4, user5] //user4 and user5 has older_userid as 2
name of user6[user4]        //user6 has older_userid as 4

role\u id=3

与旧用户相同（具有role\u id的用户始终具有旧用户id>0）

我被困在这里了。想想这样的事情：

$this->db->select('u.id, u.name, GROUP_CONCAT(old_userid) AS uid')->from('user u');
$this->db->group_by('u.old_userid');

请帮助某人…-）

我很欣赏纯sql的答案。

你可以这样做

$this->db
     ->select('o.name, GROUP_CONCAT(u.name) AS `old_users`' ,FALSE)
     ->from('user u')
     ->join('user o','u.old_userid=o.id','LEFT')
     ->where('u.role_id',3)
     ->group_by('u.old_userid')
     ->get()
     ;

---

注意：您需要在select的第二个参数中设置pas FALSE，这样它就不会受到额外的反勾号的影响。非常感谢！：-）