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
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
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
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!
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
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
