php - 查询仅从所有用户获取最后一条消息

Question

我需要编写一个查询来使用 mysql 表从两列中检索值

我的桌子有以下结构

| ID | to_user_id | from_user_id |  message  |        datetime     | 
| 1  |     21     |      07      |    hi     | 2012-05-10 04:13:01 | 
| 2  |     07     |      21      |   hello   | 2012-05-10 04:17:51 | 
| 3  |     21     |      11      | what's up | 2012-05-10 04:21:51 |
| 4  |     21     |      02      |  hi dude  | 2012-05-10 04:43:51 |

我想编写一个查询以仅从所有 from_user_id 获取最新消息

我会尝试

$query="select * from messages WHERE datetime = (SELECT MAX(datetime) FROM messages)";

但它显示了错误

'where 子句'中的未知列'datetime'

Answer 1

select t1.from_user_id, t1.message
from t1
join (
select from_user_id, max([datetime]) as [datetime]
from t1
group by from_user_id
) t2 on t1.from_user_id = t2.from_user_id
and t1.[datetime] = t2.[datetime]

Answer 2

你也可以试试：

select a.from_user_id, a.message
from table1 a
WHERE a.datetime = (SELECT MAX(datetime) FROM table1 x WHERE x.from_user_id = a.from_user_id)