Question

我有一张看起来像这样的表：

+-----+--------------+---------+----------+---------------------+---------------------+
| id  | username     | case_id | status   | date_assigned       | date_removed        |
+-----+--------------+---------+----------+---------------------+---------------------+
|  21 | hemcreynolds | 802     | inactive | 2007-08-20 07:15:48 | 2008-04-15 16:08:54 |
| 133 | licox        | 868     | active   | 2007-09-02 11:55:37 | 0000-00-00 00:00:00 |
|   3 | kashepherd   | 794     | inactive | 2007-08-17 13:54:39 | 2010-08-12 14:40:37 |
|   4 | tithomas     | 795     | active   | 2007-08-17 13:56:06 | 0000-00-00 00:00:00 |
|   5 | stcosse      | 795     | inactive | 2007-08-17 13:56:26 | 2007-09-25 10:02:03 |
|   6 | mefong yit   | 914     | active   | 2007-08-17 14:00:29 | 0000-00-00 00:00:00 |
|   7 | tamouledoux  | 842     | active   | 2007-08-17 14:09:16 | 0000-00-00 00:00:00 |
|   9 | chgross      | 903     | active   | 2007-08-18 08:38:38 | 0000-00-00 00:00:00 |
|  20 | ticrane      | 802     | inactive | 2007-08-20 07:15:41 | 2008-04-15 16:44:35 |
|  19 | hemcreynolds | 818     | inactive | 2007-08-20 07:15:12 | 2008-04-15 16:08:57 |
+-----+--------------+---------+----------+---------------------+---------------------+

我想运行查询以查找用户已按用户名分配的情况，例如

SELECT username,case_id from assignments WHERE username = 'ticrane'

然后我想找到分配给我的用户（ticrane）所分配的案例的所有其他用户：

 SELECT * FROM assignements WHERE case_id = (all of the case ids generated by the previous query)

我想在一个查询中执行此操作。我想这里的某种联接是有序的，但我认为你不能从同一个表中加入结果。困惑。有什么建议吗？

Answer 1

将表连接到自身是完全合法的，但是您需要为表的两个“副本”中的至少一个提供别名，以使该语句不含糊。

SELECT
    assignments_join.*
FROM
    assignments AS assignments_base
    JOIN assignments AS assignments_join ON
        assignments_base.case_id = assignments_join.case_id
WHERE
    assignments_base.username = 'ticrane'

Answer 2

您需要在作业表

上进行自我加入

select a.username, a.case_id
from assignments a inner join assignments b
on a.case_id = b.case_id
where b.username = 'ticrane'

Answer 3

除了Hammerite和No'am Newman所示的解决方案之外，你还可以使用一个子选择：

SELECT * 
FROM assignements 
WHERE case_id IN (SELECT case_id from assignments WHERE username = 'ticrane')

在mysql中对查询运行查询

3 个答案: