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

在这篇优秀的文章中有一个更快的CTE实现: http://weblogs.sqlteam.com/mladenp/archive/2009/07/28/SQL-Server-2005-Fast-Running-Totals.aspx

这个线程中的问题可以这样表达:

    DECLARE @RT INT
    SELECT @RT = 0

    ;
    WITH  abcd
            AS ( SELECT TOP 100 percent
                        id
                       ,SomeNumt
                       ,MySum
                       order by id
               )
      update abcd
      set @RT = MySum = @RT + SomeNumt
      output inserted.*

其他回答

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

最新版本的SQL Server(2012)允许以下。

SELECT 
    RowID, 
    Col1,
    SUM(Col1) OVER(ORDER BY RowId ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Col2
FROM tablehh
ORDER BY RowId

or

SELECT 
    GroupID, 
    RowID, 
    Col1,
    SUM(Col1) OVER(PARTITION BY GroupID ORDER BY RowId ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Col2
FROM tablehh
ORDER BY RowId

这个更快。分区版本在34秒内完成,超过500万行。

感谢Peso,他在另一个回答中提到的SQL Team线程上发表了评论。

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

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

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;