多对多表加入Pivot

kar*_*dia 9 mysql pivot many-to-many

我现在有类似的两个表users,并programs正在通过一个的方式很多一对多关系链接link表.

mysql> select * from users;
+----+----------+
| id | name     |
+----+----------+
|  1 | Jonathan |
|  2 | Little   |
|  3 | Annie    |
|  4 | Bob      |
+----+----------+
4 rows in set (0.00 sec)

mysql> select * from programs;
+----+----------------------+
| id | name                 |
+----+----------------------+
|  1 | Microsoft Word       |
|  2 | Microsoft Excel      |
|  3 | Microsoft PowerPoint |
+----+----------------------+
3 rows in set (0.00 sec)

mysql> select * from link;
+---------+------------+
| user_id | program_id |
+---------+------------+
|       1 |          1 |
|       1 |          2 |
|       1 |          3 |
|       2 |          2 |
|       3 |          1 |
|       3 |          4 |
+---------+------------+
6 rows in set (0.00 sec)
Run Code Online (Sandbox Code Playgroud)

我理解如何连接表并返回这种结果:

mysql> select users.name, programs.name from linker
    -> join users on users.id = linker.user_id
    -> join programs on programs.id = linker.program_id;
+----------+----------------------+
| name     | name                 |
+----------+----------------------+
| Jonathan | Microsoft Word       |
| Jonathan | Microsoft Excel      |
| Jonathan | Microsoft PowerPoint |
| Little   | Microsoft Excel      |
| Annie    | Microsoft Word       |
+----------+----------------------+
Run Code Online (Sandbox Code Playgroud)

但我真正想要的是更复杂一点:

+----------+-----------------------------------------------------+
| name     | name                                                |
+----------+-----------------------------------------------------+
| Jonathan | Microsoft Word,Microsoft Excel,Microsoft PowerPoint |
| Little   | Microsoft Excel                                     |
| Annie    | Microsoft Word                                      |
+----------+-----------------------------------------------------+
Run Code Online (Sandbox Code Playgroud)

我假设GROUP_CONCAT()某个地方有一个抛出的命令,但我似乎无法保持结果看起来像这样:

mysql> select users.name, group_concat(programs.name) from linker
    -> join users on users.id = linker.user_id
    -> join programs on programs.id = linker.program_id;
+----------+------------------------------------------------------------------------------------+
| name     | group_concat(programs.name)                                                        |
+----------+------------------------------------------------------------------------------------+
| Jonathan | Microsoft Word,Microsoft Excel,Microsoft PowerPoint,Microsoft Excel,Microsoft Word |
+----------+------------------------------------------------------------------------------------+
Run Code Online (Sandbox Code Playgroud)

任何人都能指出我正确的方向吗?

Rya*_*yan 10

你需要指定一个DISTINCT,即

select users.name, group_concat( DISTINCT programs.name)
Run Code Online (Sandbox Code Playgroud)

请在此处查看MySQL文档.

尝试将您的查询更改为:

SELECT users.name, group_concat(programs.name) 
from users
LEFT JOIN linker on linker.user_id = users.id
LEFT JOIN programs on linker.program_id = programs.id
GROUP BY users.id
Run Code Online (Sandbox Code Playgroud)

这将为您null提供任何没有与之关联的程序的用户.要过滤掉它们,只需添加一个WHERE programs.id IS NOT NULL.