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
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
其他回答
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。
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
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
SQL小提琴示例
输出
| ID | SOMENUMT | SUM |
-----------------------
| 1 | 10 | 10 |
| 2 | 12 | 22 |
| 3 | 3 | 25 |
| 4 | 15 | 40 |
| 5 | 23 | 63 |
编辑:这是一个通用的解决方案，可以在大多数数据库平台上工作。如果有更好的解决方案适用于你的特定平台(例如，gareth的平台)，那就使用它!
一旦创建了表-
select
A.id, A.SomeNumt, SUM(B.SomeNumt) as sum
from @t A, @t B where A.id >= B.id
group by A.id, A.SomeNumt
order by A.id
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
