ORDER BY日期和时间BEFORE在mysql中的GROUP BY名称

Question

我有一个这样的表:

name    date         time
tom | 2011-07-04 | 01:09:52
tom | 2011-07-04 | 01:09:52
mad | 2011-07-04 | 02:10:53
mad | 2009-06-03 | 00:01:01

我想要最早的名字:

SELECT * 
ORDER BY date ASC, time ASC 
GROUP BY name

( - >不起作用!)

现在它应该先让我疯狂(早期约会)然后汤姆

但是GROUP BY name ORDER BY date ASC, time ASC先给我一个新的疯狂,因为它在它排序之前就已经分组了!

再次:问题是我无法按照日期和时间排序,因为GROUP BY必须在ORDER BY之前!

Answer 1

另一种方法:

SELECT * 
FROM (
    SELECT * 
    ORDER BY date ASC, time ASC 
) AS sub
GROUP BY name

它击中第一个匹配结果的GROUP BY组.如果第一次匹配命中恰好是您想要的那个,那么一切都应该按预期工作.

我更喜欢这种方法,因为子查询具有逻辑意义,而不是将其与其他条件混淆.

Answer 2

我想这就是你要求的:

SELECT name, min(date)
FROM myTable
GROUP BY name
ORDER BY min(date)

当时,您必须通过STR_TO_DATE创建一个mysql日期:

STR_TO_DATE(date + ' ' + time, '%Y-%m-%d %h:%i:%s')

所以:

SELECT name, min(STR_TO_DATE(date + ' ' + time, '%Y-%m-%d %h:%i:%s'))
FROM myTable
GROUP BY name
ORDER BY min(STR_TO_DATE(date + ' ' + time, '%Y-%m-%d %h:%i:%s'))

Answer 3

由于我不允许对user1908688的回答发表评论,这里有一个MariaDB用户的提示:

SELECT *
FROM (
     SELECT *
     ORDER BY date ASC, time ASC
     LIMIT 18446744073709551615
     ) AS sub
GROUP BY sub.name

https://mariadb.com/kb/en/mariadb/why-is-order-by-in-a-from-subquery-ignored/

Answer 4

这对我有用:

SELECT *
FROM your_table
WHERE id IN (
    SELECT MAX(id)
    FROM your_table
    GROUP BY name
);