问候,
我对group_concat函数有一个非常棘手的问题,我想了几天没有成功,如何解决它。
有3张表:
tasks task_tag tag
+---------+----------+ +---------+----------+ +---------+----------+
| task_id | data | | task_id | tag_id | | tag_id | name |
+---------+----------+ +---------+----------+ +---------+----------+
| 1 | task 1 | | 1 | 5 | | 5 | work |
| 2 | task 2 | | 1 | 7 | | 6 | school |
| 3 | task 3 | | 2 | 6 | | 7 | home |
+---------+----------+ +---------+----------+ +---------+----------+
当需要检索所有任务时,将它们关联的所有标签作为结果列,以下查询没有问题:
SELECT t.task_id, t.data, GROUP_CONCAT(tg.name) AS tags
FROM tasks t
LEFT JOIN task_tag tt ON tt.task_id = t.task_id
LEFT JOIN tag tg ON tg.tag_id = tt.tag_id
GROUP BY t.task_id
result
+---------+-----------+-------------+
| task_id | data | tags |
+---------+-----------+-------------+
| 1 | task 1 | work,home |
| 2 | task 2 | school |
| 3 | task 3 | NULL |
+---------+-----------+-------------+
问题是当我需要从一个 EXACT 标签中选择任务时,我仍然需要保留标签列以及所有任务的关联标签。我尝试使用以下查询,但由于结果受 WHERE tag_id = 5 的限制,因此 group_concat 函数仅检索 id = 5 的标签,例如“work”:
SELECT t.task_id, t.data, GROUP_CONCAT(tg.name) AS tags
FROM tasks t
LEFT JOIN task_tag tt ON tt.task_id = t.task_id
LEFT JOIN tag tg ON tg.tag_id = tt.tag_id
WHERE tg.tag_id = 5
GROUP BY t.task_id
result
+---------+-----------+-------------+
| task_id | data | tags |
+---------+-----------+-------------+
| 1 | task 1 | work |
+---------+-----------+-------------+
我想要达到的结果是:
result
+---------+-----------+-------------+
| task_id | data | tags |
+---------+-----------+-------------+
| 1 | task 1 | work,home |
+---------+-----------+-------------+
非常感谢您提供有关如何解决此问题的任何建议!