表格中包含
等数据<--------------------------->
no category value
<--------------------------->
1 1 2000
2 1 1000
3 2 500
4 3 3000
5 1 2000
6 2 -500
7 3 5000
8 1 -1000
我希望输出像
<---------------------------------->
no category amount Sum
<---------------------------------->
1 1 2000 2000
2 1 1000 3000
3 2 500 500
4 3 3000 3000
5 1 2000 5000
6 2 -500 0
7 3 5000 8000
8 1 -1000 4000
例如:take category=1
所以,
no:1
amount=2000--->total=2000
no:2
amount=1000---->total=3000
no:5
amount=2000---->total=5000
no:8
amount=-1000---->total=4000
表示每行的每个类别的总和 并且具有更好的性能
答案 0 :(得分:0)
您需要使用行号自行加入两个表:
;WITH t(No,Category,value,row) AS
(
SELECT No, category, value
,ROW_NUMBER() OVER (Partition by category ORDER BY no) AS row
FROM Table1
)
SELECT t1.No, t1.category, t1.value, sum(t2.value) as sums
FROM t t1
JOIN t t2
ON t1.row >= t2.row
AND t1.Category = t2.Category
GROUP BY t1.No, t1.category, t1.value;
输出:
| NO | CATEGORY | VALUE | SUMS |
--------------------------------
| 1 | 1 | 2000 | 2000 |
| 2 | 1 | 1000 | 3000 |
| 3 | 2 | 500 | 500 |
| 4 | 3 | 3000 | 3000 |
| 5 | 1 | 2000 | 5000 |
| 6 | 2 | -500 | 0 |
| 7 | 3 | 5000 | 8000 |
| 8 | 1 | -1000 | 4000 |
答案 1 :(得分:0)
您需要的是计算滚动总数。 我知道的SQL Server 2005/2008 / 2008 R2中最快的方法如下所述:https://stackoverflow.com/a/13744550/1744834。您必须在每个类别中包含唯一的secuential列才能使用此方法,如下所示:
create table #t (no int, category int, value int, id int, primary key (category, id))
insert into #t (no, category, value, id)
select no, category, value, row_number() over (partition by category order by no)
from test
;with
CTE_RunningTotal
as
(
select T.no, T.category, T.value, cast(T.value as decimal(29, 10)) as total_value, T.id
from #t as T
where T.id = 1
union all
select T.no, T.category, T.value, cast(T.value + C.total_value as decimal(29, 10)) as total_value, T.id
from CTE_RunningTotal as C
inner join #t as T on T.id = C.id + 1 and T.category = C.category
)
select C.no, C.category, C.value, C.value
from CTE_RunningTotal as C
option (maxrecursion 0)
你也可以使用更短的查询,但性能更差(我认为它对于递归CTE来说是O(N ^ 2)对O(N)):
select t1.no, t1.category, t1.value, sum(t2.value)
from test as t1
inner join test as t2 on t2.category = t1.category and t2.no <= t1.no
group by t1.no, t1.category, t1.value
order by t1.no
答案 2 :(得分:-1)
您可以使用以下sql函数SQL functions它向您展示如何使用它以及如何应用它