MySQL加入，包括来自返回分隔值的所有值

Question

我正在使用表格。为了说明问题：

mysql> SELECT users.id, username, REPLACE(selected_groups, '\r', ', '), groups.id, groups.label FROM users LEFT JOIN groups ON users.selected_groups; 
+----+------------------------+--------------------------------------+------+---------------+ 
| id | username    | REPLACE(selected_groups, '\r', ', ') | id | label   | 
+----+------------------------+--------------------------------------+------+---------------+ 
| 7 | Greg     | 1, 2, 3,        | 1 | Administrator | 
| 7 | Greg     | 1, 2, 3,        | 2 | Member  | 
| 7 | Greg     | 1, 2, 3,        | 3 | Banned  | 
| 8 | Steve     | 1, 2,        | 1 | Administrator | 
| 8 | Steve     | 1, 2,        | 2 | Member  | 
| 8 | Steve     | 1, 2,        | 3 | Banned  | 
| 9 | Steve The Banned Admin | 1, 3,        | 1 | Administrator | 
| 9 | Steve The Banned Admin | 1, 3,        | 2 | Member  | 
| 9 | Steve The Banned Admin | 1, 3,        | 3 | Banned  | 
| 10 | Example man   | 1, 2, 3,        | 1 | Administrator | 
| 10 | Example man   | 1, 2, 3,        | 2 | Member  | 
| 10 | Example man   | 1, 2, 3,        | 3 | Banned  | 
+----+------------------------+--------------------------------------+------+---------------+ 
12 rows in set (0.00 sec) 

正如你所看到的，我有一个用户表和一个组表。在用户表内是一个'selected_groups'列，其中相关组使用返回分隔值（'\ r'）来定义。

问题是我不太确定我如何在标签字段中返回每个必需的标签。所产生的输出在理想情况下是这样的：

+----+------------------------+--------------------------------------+------+-------------------------------+ 
| id | username    | REPLACE(selected_groups, '\r', ', ') | id | groups.label     | 
+----+------------------------+--------------------------------------+------+-------------------------------+ 
| 7 | Greg     | 1, 2, 3,        | 1 | Administrator, Member, Banned | 
| 8 | Steve     | 1, 2,        | 1 | Administrator, Member   | 
| 9 | Steve The Banned Admin | 1, 3,        | 1 | Administrator, Banned   | 
| 10 | Example man   | 1, 2, 3,        | 1 | Administrator, Member, Banned | 
+----+------------------------+--------------------------------------+------+-------------------------------+ 

你认为这是可能的，如果是这样，没有任何人有任何提示/我的解决方案？

非常感谢， 瑞奇

Answer 1

你应该看看group_concat。

对于用户标识和用户名以及group_concat标签上的group by子句，它应该做到这一点。

我想是这样的：

SELECT users.id, username, REPLACE(selected_groups, '\r', ', '), GROUP_CONCAT(DISTINCT groups.label) 
FROM users 
LEFT JOIN groups ON users.selected_groups; 
GROUP BY users.id, username 

（这不是测试，所以也许这将需要一些修改）

编辑：我删除从查询中groups.id，因为它没有任何意义。