我有两个看起来像这样的表:
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,Table A
包含用户uID
的属性数据,Table B
包含用户uID
的权限数据。
此刻,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以获取:
+-------+------------+----------+
| attrs | values | perms |
+-------+------------+----------+
| fn,ln | John,Smith | admin |
+-------+------------+----------+
| fn,fn | Jim,Bean | news,cms |
+-------+------------+----------+
答案 0 :(得分:4)
GROUP_CONCAT
需要其他参数,如其文档页面here中所述。
你想要的是distinct
:
SELECT GROUP_CONCAT(distinct `a`.`attr`) AS `attrs` . . .