我有两个表格,如下所示:
表A:
+-----+-----+------+-------+
| aID | uID | attr | value |
+-----+-----+------+-------+
| 1 | 1 | fn | john |
+-----+-----+------+-------+
| 2 | 1 | ln | smith |
+-----+-----+------+-------+
| 3 | 2 | fn | jim |
+-----+-----+------+-------+
| 4 | 2 | ln | bean |
+-----+-----+------+-------+
表 B:
+-----+-----+-------+-------+
| bID | uID | perm | value |
+-----+-----+-------+-------+
| 1 | 1 | admin | 1 |
+-----+-----+-------+-------+
| 2 | 2 | news | 1 |
+-----+-----+-------+-------+
| 3 | 2 | cms | 1 |
+-----+-----+-------+-------+
如图所示,表A
保存了用户uID
的属性数据,表B
保存了用户的权限数据uID
.
目前,我正在使用:
SELECT GROUP_CONCAT(`a`.`attr`) AS `attrs`
, GROUP_CONCAT(`a`.`value`) AS `values`
, GROUP_CONCAT(`b`.`perm`) AS `perms`
FROM `a`
JOIN `b`
ON `a`.`uID` = `b`.`uID`
GROUP BY `a`.`uID`, `b`.`uID`
但它给了我一个结果:
+-------------+-------------------+-------------------+
| attrs | values | perms |
+-------------+-------------------+-------------------+
| fn,ln | John,Smith | admin,admin |
+-------------+-------------------+-------------------+
| fn,fn,ln,ln | Jim,Jim,Bean,Bean | news,cms,news,cms |
+-------------+-------------------+-------------------+
我需要在查询中更改什么才能获得:
+-------+------------+----------+
| attrs | values | perms |
+-------+------------+----------+
| fn,ln | John,Smith | admin |
+-------+------------+----------+
| fn,fn | Jim,Bean | news,cms |
+-------+------------+----------+
最佳答案
GROUP_CONCAT
接受额外的参数,如其文档页面 here 中所述.
你想要的是distinct
:
SELECT GROUP_CONCAT(distinct `a`.`attr`) AS `attrs` . . .
关于mysql - 修复连接的多对多 MySQL 查询,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13958788/