Question

我需要在一个表中为一个特定用户选择最新的三个条目，并使用另一个列的唯一值。这是我的数据库的结构：

time_tracking
    id
    user
    todo
    tracked_time
    start_date
    start_time
    running

因此todo列必须是唯一的，并且我需要按user和start_date降序排序的每个start_time最近的三个。

我试图在Google周围使用DISTINCT和许多其他答案，但似乎都没有帮助使用像

这样的东西

SELECT * FROM time_tracking WHERE user = 4735350 GROUP BY todo ORDER BY start_date DESC, start_time DESC LIMIT 3

ALMOST 获取它，但如果它是以前的值，它会跳过一些todos，如下所示：

id  |user    |todo    |start_date    |start_time
1   |4735350 |1       |2014-09-10    |10:00:00
2   |4735350 |2       |2014-09-11    |10:00:00
3   |4735350 |1       |2014-09-11    |11:00:00 -- gets skipped because of id=1 even though it has a more recent date

编辑：我需要表格中的所有列，因此我似乎无法使用DISTINCT。

Answer 1

您可以通过操作获得最大的start_date值，并可以使用EXISTS

SQL FIDDLE

SELECT * FROM 
time_tracking t
WHERE user = 4735350 
AND exists
(
SELECT  1
FROM time_tracking 
WHERE user = 4735350 
GROUP BY todo 
HAVING MAX(start_date) = t.start_date
      and MAX(start_time) = t.start_time
)
order by start_date desc, start_time desc
LIMIT 3

Answer 2

你可以试试这个。

SELECT * FROM time_tracking WHERE user = 4735350 GROUP BY 1 ORDER BY start_date DESC, start_time DESC LIMIT 3

将group by从待办事项更改为1

编辑：好吧，我正在尝试一个新的解决方案，但我得去，但至少，尝试这样的事情：

select t1.* from (select todo, MAX(start_date) start_date from time_tracking group by todo ) t2 join time_tracking t1 on t1.todo = t2.todo and t1.start_date = t2.start_date order by start_date desc, start_time desc LIMIT 3;

问候。

Answer 3

SELECT [user],todo, max(start_time),max(start_date) FROM time_tracking WHERE [user] = 4735350 GROUP BY [user],todo

尝试此查询...工作

GROUP BY不遵循ORDER BY

3 个答案: