示例数据:
create table #temp
(
dateRef date,
a float,
b float,
c float
)
insert into #temp (dateRef, a, b, c) values ('20050101',1000,500,0)
insert into #temp (dateRef, a, b, c) values ('20050201',1000,1000,0)
insert into #temp (dateRef, a, b, c) values ('20050301',4000,4000,3000)
insert into #temp (dateRef, a, b, c) values ('20050401',2000,2000,1000)
insert into #temp (dateRef, a, b, c) values ('20050501',1000,2000,0)
insert into #temp (dateRef, a, b, c) values ('20050601',2000,2000,0)
insert into #temp (dateRef, a, b, c) values ('20050701',2000,2000,1000)
insert into #temp (dateRef, a, b, c) values ('20050801',2000,2000,1000)
---
insert into #temp (dateRef, a, b, c) values ('20060301',1000,1000,0)
insert into #temp (dateRef, a, b, c) values ('20060601',1000,1000,0)
insert into #temp (dateRef, a, b, c) values ('20060701',2000,2000,0)
insert into #temp (dateRef, a, b, c) values ('20060801',2000,2000,1000)
---
insert into #temp (dateRef, a, b, c) values ('20070101',1000,1000,0)
获得此预期结果的最佳方法是什么? (使用SQL Server 2012)
DateRef a b c
----------------------------------------
--- 20050301 6000 5500 3000
--- 20050401 2000 2000 1000
--- 20050701 5000 6000 1000
--- 20050801 2000 2000 1000
--- 20060801 6000 6000 1000
逻辑: 当C> 0,例如:当data ='20050301',c = 3000(c> 0)时,如果日期是连续的,则需要总和(a)(在这种情况下,它们是,当数据= 20050701 c = 1000(c> 0)时,如果日期是连续的(在这种情况下它们是,20050501-20050601-20050701),则数据= 20050701 c = 1000(c> 0),当date ='20050801'c时= 1000(c> 0)然后我需要总和(a)在这种情况下只有年20050801等等
答案 0 :(得分:1)
尝试此查询:
select
dateRef = max(dateRef), a = sum(a), b = sum(b), c = sum(c)
from (
select
*, rn = datediff(mm, '19000101', dateRef) - row_number() over (order by dateRef)
, grp = isnull(sum(iif(c > 0, 1, 0)) over (order by dateRef rows between unbounded preceding and 1 preceding), 0)
from
#temp
) t
group by rn, grp
having sum(c) > 0
在查询中使用两列进行分组。 rn
- 查找连续的行grp
- 对行进行分组,其中c> 0,前面的行,其中c = 0。
输出
dateRef a b c
---------------------------------
2005-03-01 6000 5500 3000
2005-04-01 2000 2000 1000
2005-07-01 5000 6000 1000
2005-08-01 2000 2000 1000
2006-08-01 5000 5000 1000
答案 1 :(得分:0)
对于Grouping
年的数据,我们可以尝试一下 -
select
year(dateref) as year,
sum(a) as a,
sum(b) as b,
sum(c) as c
from #temp
group by year(dateref)
输出 -
year a b c
2005 15000 15500 6000
2006 6000 6000 1000
2007 1000 1000 0