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
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;
其他回答
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
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.*
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
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。
