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