SQL Server的增量爲每個新的獨特的價值

Question

我試圖得到像下面這樣的結果的行號：

1 foo 1-Jan 
2 bar 2-Feb 
3 baz 3-Mar 
3 baz 4-Apr 
3 baz 5-May 
4 quz 6-Jun 
4 quz 7-Jul 

帶T-SQL：

SELECT ROW_NUMBER() OVER (PARTITION BY [ID] ORDER BY [DateTime]) AS [SetNum], * 
FROM [ComplicatedQuery] 

但不是我越來越：

1 foo 1-Jan 
1 bar 2-Feb 
1 baz 3-Mar 
2 baz 4-Apr 
3 baz 5-May 
1 quz 6-Jun 
2 quz 7-Jul 

我怎樣才能改變它遞增行/集數爲[ID]每個新的獨特的價值？

Answer 1

我會用的DENSE_RANK()代替ROW_NUMBER()這樣的：

DENSE_RANK() OVER(ORDER BY [Column_with_foo_bar_baz_coco_jambo_etc]) AS Rnk 

提示：看例子A. Ranking rows within a partition從documentation：

Answer 2

我們需要使用兩個窗口聚集以達到您想要的結果 - 第一確保爲同一ID的所有行分配一個值，然後在排序標準中使用：

declare @t table (Desired int,ID varchar(9),Dt datetime) 
insert into @t(Desired,ID,Dt) values 
(1,'foo','20170101'), 
(2,'bar','20170202'), 
(3,'baz','20170303'), 
(3,'baz','20170404'), 
(3,'baz','20170505'), 
(4,'quz','20170606'), 
(4,'quz','20170707') 

;With MinDt as (
select *,MIN(Dt) OVER (PARTITION BY ID) as MinDt 
from @t 
) 
select *,DENSE_RANK() OVER (ORDER BY MinDt,ID) as rk 
from MinDt 

個

結果：

Desired  ID  Dt      MinDt     rk 
----------- --------- ----------------------- ----------------------- -------------------- 
1   foo  2017-01-01 00:00:00.000 2017-01-01 00:00:00.000 1 
2   bar  2017-02-02 00:00:00.000 2017-02-02 00:00:00.000 2 
3   baz  2017-03-03 00:00:00.000 2017-03-03 00:00:00.000 3 
3   baz  2017-04-04 00:00:00.000 2017-03-03 00:00:00.000 3 
3   baz  2017-05-05 00:00:00.000 2017-03-03 00:00:00.000 3 
4   quz  2017-06-06 00:00:00.000 2017-06-06 00:00:00.000 4 
4   quz  2017-07-07 00:00:00.000 2017-06-06 00:00:00.000 4 

的ID在DENSE_RANK列入是作爲在殼體的多個仲裁器ID值具有相同的最小Dt值。

Answer 3

SELECT Distinct Dense_Rank() OVER (ORDER BY ID) AS [SetNum], ID into #temp 
FROM [ComplicatedQuery] 


SELECT t.SetNum AS [SetNum], cq.* 
FROM [ComplicatedQuery] cq join #temp t on cq.ID=t.ID order by t.SetNum