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:
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
其他回答
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
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
对于SQL Server 2012以后，它可以很容易:
SELECT id, SomeNumt, sum(SomeNumt) OVER (ORDER BY id) as CumSrome FROM @t
因为SUM的ORDER BY子句默认表示窗口框的前一行和当前行范围为无界(“一般备注”在https://msdn.microsoft.com/en-us/library/ms189461.aspx)
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
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。
