Mysql 获取每个对话的收件人id
我有一张这样的桌子Mysql 获取每个对话的收件人id,mysql,sql,Mysql,Sql,我有一张这样的桌子 | id-AI | from | to | text |created_date +----+---------+-------+-------------+------------- | 1 | 7 | 1 | from 7 to 1 |2015-11-06 04:59:14 | 2 | 1 | 2 | from 1 to 2 |2015-11-06 04:57:06 | 3 | 7
| id-AI | from | to | text |created_date
+----+---------+-------+-------------+-------------
| 1 | 7 | 1 | from 7 to 1 |2015-11-06 04:59:14
| 2 | 1 | 2 | from 1 to 2 |2015-11-06 04:57:06
| 3 | 7 | 1 | from 7 to 1 |2015-11-06 04:25:45
| 4 | 1 | 6 | from 1 to 6 |2015-11-06 04:25:32
| 5 | 1 | 5 | from 1 to 5 |2015-11-06 04:25:16
我想要一个这样的列表。一个打击我或我打击他/她或我们进行了长时间对话的用户列表。我想要的列表应该是最新的顺序,就像facebook消息一样
| users |
+-------+
| 7
| 2
| 6
| 5
1是主要用户id。获取1所涉及的列表(在收件人或发件人中),然后这称为对话。您可以使用以下查询:
SELECT IF (`from` = 1, `to`, `from`)
FROM mytable
WHERE 1 IN (`from`, `to`)
ORDER BY created_date DESC
这将返回包含user\u id=1
的每个对话的创建者或接收者
获取一组不同的用户id
值的有序集合使用此查询:
SELECT IF (`from` = 1, `to`, `from`)
FROM mytable
WHERE 1 IN (`from`, `to`)
GROUP BY IF (`from` = 1, `to`, `from`)
ORDER BY MAX(created_date) DESC
使用联合体:
SELECT user
FROM (SELECT to AS user, created_date
FROM mytable
WHERE from = 1)
UNION
SELECT from AS user, created_date
FROM mytable
WHERE to = 1) x
GROUP BY user
ORDER BY MAX(created_date) DESC
有什么问题吗?编写一个查询,在
from
中找到他,另一个查询在to
中找到他,并将它们与UNION
组合。是否按日期排列最新顺序?如果使用按创建的顺序_datedesc
,则会出现这种情况。请给出sql查询。我尝试使用UNION和ORDER BY,但结果不是我想要的。谢谢。它返回了多个值。我想要一个清晰的用户列表。显然,在“收件人”或“发件人”字段中有多个用户id条目,因为这是对话。@拉吉布霍森看一下我所做的编辑。