ps.*_*ps. 165 sql-server sql-server-2008
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
上面的选择返回以下内容.
id SomeNumt
1 10
2 12
3 3
4 15
5 23
我如何得到以下内容
id srome CumSrome
1 10 10
2 12 22
3 3 25
4 15 40
5 23 63
Red*_*ter 203
select t1.id, t1.SomeNumt, SUM(t2.SomeNumt) as sum
from @t t1
inner join @t t2 on t1.id >= t2.id
group by t1.id, t1.SomeNumt
order by t1.id
产量
| ID | SOMENUMT | SUM |
-----------------------
| 1 | 10 | 10 |
| 2 | 12 | 22 |
| 3 | 3 | 25 |
| 4 | 15 | 40 |
| 5 | 23 | 63 |
编辑:这是一个适用于大多数数据库平台的通用解决方案.如果您的特定平台有更好的解决方案(例如,gareth),请使用它!
小智 175
最新版本的SQL Server(2012)允许以下内容.
SELECT
RowID,
Col1,
SUM(Col1) OVER(ORDER BY RowId ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Col2
FROM tablehh
ORDER BY RowId
要么
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
这甚至更快.对于我来说,分区版本在34秒内完成超过500万行.
感谢Peso,他评论了另一个答案中提到的SQL Team线程.
小智 22
对于SQL Server 2012以上,它可能很简单:
SELECT id, SomeNumt, sum(SomeNumt) OVER (ORDER BY id) as CumSrome FROM @t
因为默认情况下,ORDER BY子句SUM意味着RANGE UNBOUNDED PRECEDING AND CURRENT ROW窗口框架(https://msdn.microsoft.com/en-us/library/ms189461.aspx上的 "一般说明" )
Dam*_*vic 12
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
Nee*_*rma 11
让我们首先创建一个带有虚拟数据的表 - >
Create Table CUMULATIVESUM (id tinyint , SomeValue tinyint)
**Now let put some data in the table**
Insert Into CUMULATIVESUM
Select 1, 10 union
Select 2, 2 union
Select 3, 6 union
Select 4, 10
在这里我加入同桌(SELF加入)
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
在这里我们现在只是总结t2的Somevalue,我们将得到ans
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
FOR SQL SERVER 2012及更高版本(更好的表现)
Select c1.ID, c1.SomeValue,
SUM (SomeValue) OVER (ORDER BY c1.ID )
From CumulativeSum c1
Order By c1.id Asc
期望的结果
ID SomeValue CumlativeSumValue
1 10 10
2 2 12
3 6 18
4 10 28
Drop Table CumulativeSum
清除虚拟表
小智 5
Select
*,
(Select Sum(SOMENUMT)
From @t S
Where S.id <= M.id)
From @t M
迟到的答案但显示出更多的可能性......
可以使用CROSS APPLY逻辑更优化累积和生成.
在分析实际查询计划时,工作效果优于INNER JOIN&OVER Clause
/* Create table & populate data */
IF OBJECT_ID('tempdb..#TMP') IS NOT NULL
DROP TABLE #TMP
SELECT * INTO #TMP
FROM (
SELECT 1 AS id
UNION
SELECT 2 AS id
UNION
SELECT 3 AS id
UNION
SELECT 4 AS id
UNION
SELECT 5 AS id
) Tab
/* Using CROSS APPLY
Query cost relative to the batch 17%
*/
SELECT T1.id,
T2.CumSum
FROM #TMP T1
CROSS APPLY (
SELECT SUM(T2.id) AS CumSum
FROM #TMP T2
WHERE T1.id >= T2.id
) T2
/* Using INNER JOIN
Query cost relative to the batch 46%
*/
SELECT T1.id,
SUM(T2.id) CumSum
FROM #TMP T1
INNER JOIN #TMP T2
ON T1.id > = T2.id
GROUP BY T1.id
/* Using OVER clause
Query cost relative to the batch 37%
*/
SELECT T1.id,
SUM(T1.id) OVER( PARTITION BY id)
FROM #TMP T1
Output:-
id CumSum
------- -------
1 1
2 3
3 6
4 10
5 15