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

让我们先用虚拟数据创建一个表:

Create Table CUMULATIVESUM (id tinyint , SomeValue tinyint)

现在让我们向表中插入一些数据;

Insert Into CUMULATIVESUM
    Select 1, 10 union 
    Select 2, 2  union
    Select 3, 6  union
    Select 4, 10 

这里我在连接同一个表(自连接)

Select c1.ID, c1.SomeValue, c2.SomeValue
From CumulativeSum c1, CumulativeSum c2
Where c1.id >= c2.ID
Order By c1.id Asc

结果:

ID  SomeValue   SomeValue
-------------------------
1   10          10
2   2           10
2   2            2
3   6           10
3   6            2
3   6            6
4   10          10
4   10           2
4   10           6
4   10          10

现在我们把t2的somvalue相加,我们就会得到答案

Select c1.ID, c1.SomeValue, Sum(c2.SomeValue) CumulativeSumValue
From CumulativeSum c1,  CumulativeSum c2
Where c1.id >= c2.ID
Group By c1.ID, c1.SomeValue
Order By c1.id Asc

对于SQL Server 2012及以上版本(性能更好):

Select 
    c1.ID, c1.SomeValue, 
    Sum (SomeValue) Over (Order By c1.ID )
From CumulativeSum c1
Order By c1.id Asc

预期的结果:

ID  SomeValue   CumlativeSumValue
---------------------------------
1   10          10
2   2           12
3   6           18
4   10          28

Drop Table CumulativeSum

其他回答

Select 
    *, 
    (Select Sum(SOMENUMT) 
     From @t S 
     Where S.id <= M.id)
From @t M

在上面(Pre-SQL12)我们看到了这样的例子:-

SELECT
    T1.id, SUM(T2.id) AS CumSum
FROM 
    #TMP T1
    JOIN #TMP T2 ON T2.id < = T1.id
GROUP BY
    T1.id

更高效的…

SELECT
    T1.id, SUM(T2.id) + T1.id AS CumSum
FROM 
    #TMP T1
    JOIN #TMP T2 ON T2.id < T1.id
GROUP BY
    T1.id

例如:如果你有一个有两列的表,一列是ID,第二列是number,并且想要找出累积和。

SELECT ID,Number,SUM(Number)OVER(ORDER BY ID) FROM T

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

试试这个

select 
    t.id,
    t.SomeNumt, 
    sum(t.SomeNumt) Over (Order by t.id asc Rows Between Unbounded Preceding and Current Row) as cum
from 
    @t t 
group by
    t.id,
    t.SomeNumt
order by
    t.id asc;