[英]SQL Group with Min and Max
I have a super simple query using group by and I just can't figure out how to get the desired result.我有一个使用 group by 的超级简单查询,但我不知道如何获得所需的结果。 It's literally a simple query with min() and max().
这实际上是一个带有 min() 和 max() 的简单查询。 I have a table where assets belong to a certain location with a date in/out (it also has multiple in/out dates without changing location), but if an asset moves back to location where its previously already been, the grouping doesn't work.
我有一个表,其中资产属于某个位置,有一个日期进/出(它也有多个进/出日期而不改变位置),但是如果资产移回之前已经存在的位置,则分组不会工作。 I tried using a combination of over(partition by...), just can't solve it.
我尝试使用over(partition by...)的组合,就是无法解决。
Table:桌子:
Asset Location Date In Date Out
------------------------------------------
00001 A 01/01/2020 13/01/2020
00001 A 14/01/2020 26/01/2020
00001 A 27/01/2020 08/02/2020
00001 B 09/02/2020 21/02/2020
00001 B 22/02/2020 05/03/2020
00001 B 06/03/2020 18/03/2020
00001 A 19/03/2020 31/03/2020
00001 A 01/04/2020 13/04/2020
00001 A 14/04/2020 26/04/2020
00001 A 27/04/2020 09/05/2020
00001 A 10/05/2020 16/09/2020
Desired result:想要的结果:
Asset Location Date In Date Out
------------------------------------------
00001 A 01/01/2020 08/02/2020
00001 B 09/02/2020 18/03/2020
00001 A 19/03/2020 16/09/2020
Actual result:实际结果:
Asset Location Date In Date Out
------------------------------------------
00001 A 01/01/2020 16/09/2020
00001 B 09/02/2020 18/03/2020
This is a groups-and-islands problem.这是一个群体和岛屿问题。 A simple solution is the difference of row numbers:
一个简单的解决方案是行号的差异:
select asset, location, min(date_in), max(date_out)
from (select t.*,
row_number() over (partition by asset order by date_in) as seqnum
row_number() over (partition by asset, location order by date_in) as seqnum_2
from t
) t
group by asset, location, (seqnum - seqnum_2);
Why this works is a little tricky to explain.为什么这行得通有点难以解释。 If you look at the subquery you'll see how the difference in row numbers defines the consecutive rows you want.
如果您查看子查询,您将看到行号的差异如何定义您想要的连续行。
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