Question

鉴于下表：

code    date1               date2
393517  07/07/2016 00:00    08/07/2016 00:00
393517  07/07/2016 00:00    11/07/2016 00:00
393517  07/07/2016 00:00    12/07/2016 00:00
393517  07/07/2016 00:00    13/07/2016 00:00
393517  08/07/2016 00:00    11/07/2016 00:00
393517  08/07/2016 00:00    12/07/2016 00:00
393517  08/07/2016 00:00    13/07/2016 00:00
393517  11/07/2016 00:00    12/07/2016 00:00
393517  11/07/2016 00:00    13/07/2016 00:00
393516  07/07/2016 00:00    08/07/2016 00:00
393516  07/07/2016 00:00    11/07/2016 00:00
393516  07/07/2016 00:00    12/07/2016 00:00
393516  07/07/2016 00:00    13/07/2016 00:00
393516  08/07/2016 00:00    11/07/2016 00:00
393516  08/07/2016 00:00    12/07/2016 00:00
393516  08/07/2016 00:00    13/07/2016 00:00
393516  11/07/2016 00:00    12/07/2016 00:00
393516  11/07/2016 00:00    13/07/2016 00:00

如何为每个date1仅选择一行，其中date2是最近的日期？

所以最终的输出是：

code    date1               date2
393517  07/07/2016 00:00    13/07/2016 00:00
393517  08/07/2016 00:00    13/07/2016 00:00
393517  11/07/2016 00:00    13/07/2016 00:00
393516  07/07/2016 00:00    13/07/2016 00:00
393516  08/07/2016 00:00    13/07/2016 00:00
393516  11/07/2016 00:00    13/07/2016 00:00

换句话说，我只为每个date1和代码保留date2的最高值。

Answer 1

试

SELECT t1.*
FROM Table1 t1
WHERE t1.date2 = (SELECT MAX(t2.date2)
             FROM Table1 t2
             WHERE t2.date1 = t1.date1 AND t2.code=t1.code)

Answer 2

听起来你只想要一个简单的GROUP BY查询：

SELECT code, date1, MAX(date2)
FROM yourTable
GROUP BY code, date1

Answer 3

select code, date1, max(date2) as date2 from table group by code, date1

MySQL选择不同的日期查询

3 个答案: