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

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。

其他回答

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.*

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。

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

在不使用任何类型的JOIN的情况下,通过使用follow查询获取一个人的累计工资:

SELECT * , (
  SELECT SUM( salary ) 
  FROM  `abc` AS table1
  WHERE table1.ID <=  `abc`.ID
    AND table1.name =  `abc`.Name
) AS cum
FROM  `abc` 
ORDER BY Name

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;