declare  @t table
    (
        id int,
        SomeNumt int
    )

insert into @t
select 1,10
union
select 2,12
union
select 3,3
union
select 4,15
union
select 5,23


select * from @t

上面的选择返回如下内容。

id  SomeNumt
1   10
2   12
3   3
4   15
5   23

我如何得到以下:

id  srome   CumSrome
1   10  10
2   12  22
3   3   25
4   15  40
5   23  63

当前回答

SQL解决方案结合“无界前行和当前行之间的行”和“和”做的正是我想要实现的。 非常感谢!

如果这能帮到谁,这是我的案子。我想在一列中累积+1,每当发现一个maker为“Some maker”(示例)。如果不是,则不增加,但显示之前的增加结果。

这段SQL:

SUM( CASE [rmaker] WHEN 'Some Maker' THEN  1 ELSE 0 END) 
OVER 
(PARTITION BY UserID ORDER BY UserID,[rrank] ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Cumul_CNT

让我得到这样的东西:

User 1  Rank1   MakerA      0  
User 1  Rank2   MakerB      0  
User 1  Rank3   Some Maker  1  
User 1  Rank4   Some Maker  2  
User 1  Rank5   MakerC      2
User 1  Rank6   Some Maker  3  
User 2  Rank1   MakerA      0  
User 2  Rank2   SomeMaker   1  

上面的解释:它从0开始计数“some maker”,some maker被找到,我们做+1。对于用户1,MakerC被找到,所以我们不做+1,而是一些制造商的垂直计数被固定为2,直到下一行。 分区是按用户划分的,所以当我们改变用户时,累积计数返回零。

我在工作,我不希望这个答案有任何优点,只是说谢谢,并以身作则,以防有人处于同样的情况。我试图结合SUM和PARTITION,但惊人的语法“无界前行和当前行之间的行”完成了任务。

谢谢! Groaker

其他回答

一个CTE版本,只是为了好玩:

;
WITH  abcd
        AS ( SELECT id
                   ,SomeNumt
                   ,SomeNumt AS MySum
             FROM   @t
             WHERE  id = 1
             UNION ALL
             SELECT t.id
                   ,t.SomeNumt
                   ,t.SomeNumt + a.MySum AS MySum
             FROM   @t AS t
                    JOIN abcd AS a ON a.id = t.id - 1
           )
  SELECT  *  FROM    abcd
OPTION  ( MAXRECURSION 1000 ) -- limit recursion here, or 0 for no limit.

返回:

id          SomeNumt    MySum
----------- ----------- -----------
1           10          10
2           12          22
3           3           25
4           15          40
5           23          63

一旦创建了表-

select 
    A.id, A.SomeNumt, SUM(B.SomeNumt) as sum
    from @t A, @t B where A.id >= B.id
    group by A.id, A.SomeNumt

order by A.id

对于SQL Server 2012以后,它可以很容易:

SELECT id, SomeNumt, sum(SomeNumt) OVER (ORDER BY id) as CumSrome FROM @t

因为SUM的ORDER BY子句默认表示窗口框的前一行和当前行范围为无界(“一般备注”在https://msdn.microsoft.com/en-us/library/ms189461.aspx)

试试这个:

CREATE TABLE #t(
 [name] varchar NULL,
 [val] [int] NULL,
 [ID] [int] NULL
) ON [PRIMARY]

insert into #t (id,name,val) values
 (1,'A',10), (2,'B',20), (3,'C',30)

select t1.id, t1.val, SUM(t2.val) as cumSum
 from #t t1 inner join #t t2 on t1.id >= t2.id
 group by t1.id, t1.val order by t1.id

你可以使用这个简单的查询进行累进计算:

select 
   id
  ,SomeNumt
  ,sum(SomeNumt) over(order by id ROWS between UNBOUNDED PRECEDING and CURRENT ROW) as CumSrome
from @t