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)
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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 |
+----------+----------------------+
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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 |
+----------+-----------------------------------------------------+
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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 |
+----------+------------------------------------------------------------------------------------+
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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)
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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
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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.
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