Question

我正在使用桌子。为了说明的话：

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)

如您所见，我有一个用户表和一个组表。在users表中有一个'selected_groups'列，其中使用返回的分隔值（'\ r'）定义相关的组。

问题是我不能确定如何在标签字段中返回每个必需的标签。结果输出最好是这样的：

+----+------------------------+--------------------------------------+------+-------------------------------+
| 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，如果是的话，有没有人对我有任何提示/解决方案？

非常感谢，瑞奇

Answer 1

您应该查看group_concat。

在用户ID和用户名上加group by子句，在标签上加group_concat，就可以了。

我在想：

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

（这未经过测试，因此可能需要进行一些修改）

编辑：我从查询中删除了groups.id，因为它没有任何意义。

MySQL Join，包括返回分隔值的所有值

1 个答案: