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
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!
其他回答
在不使用任何类型的JOIN的情况下，通过使用follow查询获取一个人的累计工资:
SELECT * , (
SELECT SUM( salary )
FROM `abc` AS table1
WHERE table1.ID <= `abc`.ID
AND table1.name = `abc`.Name
) AS cum
FROM `abc`
ORDER BY Name
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
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)
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
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
