Question

 rN         rD          rnc      d     expectedResult
abc1m      2010-03-31   abc     5.7    5.7 + 1.7 +9.6
abc3m      2010-03-31   abc     5.7    5.7 + 1.7 +9.6
abc1y      2010-03-31   abc     5.7    5.7 + 1.7 +9.6
xfx1m      2010-03-31   xfx     1.7    5.7 + 1.7 +9.6
xfx3m      2010-03-31   xfx     1.7    5.7 + 1.7 +9.6
xfx1y      2010-03-31   xfx     1.7    5.7 + 1.7 +9.6
tnt1m      2010-03-31   tnt     9.6    5.7 + 1.7 +9.6
tnt3m      2010-03-31   tnt     9.6    5.7 + 1.7 +9.6
tnt1y      2010-03-31   tnt     9.6    5.7 + 1.7 +9.6
------------------------------------
abc1m      2010-04-01   abc     2.2    2.2 + 8.9 + 5.5
abc3m      2010-04-01   abc     2.2    2.2 + 8.9 + 5.5
abc1y      2010-04-01   abc     2.2    2.2 + 8.9 + 5.5
xfx1m      2010-04-01   xfx     8.9    2.2 + 8.9 + 5.5
xfx3m      2010-04-01   xfx     8.9    2.2 + 8.9 + 5.5
xfx1y      2010-04-01   xfx     8.9    2.2 + 8.9 + 5.5
tnt1m      2010-04-01   tnt     5.5    2.2 + 8.9 + 5.5
tnt3m      2010-04-01   tnt     5.5    2.2 + 8.9 + 5.5
tnt1y      2010-04-01   tnt     5.5    2.2 + 8.9 + 5.5

预期结果是特定日期的不同rnc的总和如何实现这一目标。我想使用类似下面的代码，但不起作用。

select *,
sum (d) over (partition by rD, distinct rnc) as expectedResult
from myTable
where ...--some condition
order by ...--order by some columns

使用SQL Server 2012，谢谢

编辑：关于暂停的问题，这是怎么回事。如果一个人只看expectedResult栏，那么它是不是很清楚？我应该添加什么才能让它变得更好？ - 每个人都有d。假设每个集合都是示例中给出的形式。（回答一条评论）

Answer 1

这里我们使用cte对相同的行进行分组。这样，我们可以只选择select中每个组的第一行。

;WITH cte 
AS
(
    SELECT  *,
            GroupRowIndex = ROW_NUMBER() OVER (PARTITION BY rateDate, rnc, d ORDER BY (SELECT 1))
    FROM myTable
)

SELECT  *,
        expectedResult = SUM(d) OVER (PARTITION BY rateDate)
FROM cte
WHERE GroupRowIndex = 1
AND ...--some condition
ORDER BY ...--order by some columns

Answer 2

由于第一列的最后2个字符是可重复的，并且您实际上是在分区中进行求和，请试一试并告诉我这是否是您要求的

create table #TempTable (rn nvarchar(10), rD date, rnc nvarchar(10), d decimal(5,2))

insert into #TempTable (rn, rD, rnc, d)
values
('abc1m','2010-03-31','abc',     5.7),
('abc3m','2010-03-31','abc',     5.7),
('abc1y','2010-03-31','abc',     5.7),
('xfx1m','2010-03-31','xfx',     1.7),
('xfx3m','2010-03-31','xfx',     1.7),
('xfx1y','2010-03-31','xfx',     1.7),
('tnt1m','2010-03-31','tnt',     9.6),
('tnt3m','2010-03-31','tnt',     9.6),
('tnt1y','2010-03-31','tnt',     9.6),
------------------------------------
('abc1m','2010-04-01','abc',     2.2),
('abc3m','2010-04-01','abc',     2.2),
('abc1y','2010-04-01','abc',     2.2),
('xfx1m','2010-04-01','xfx',     8.9),
('xfx3m','2010-04-01','xfx',     8.9),
('xfx1y','2010-04-01','xfx',     8.9),
('tnt1m','2010-04-01','tnt',     5.5),
('tnt3m','2010-04-01','tnt',     5.5),
('tnt1y','2010-04-01','tnt',     5.5)

select rn, rD, rnc, d, SUM(d) over (partition by right(rn,2), rD) as 'Sum' 
from #TempTable
order by Rd

Answer 3

您可以使用交叉申请来应用基于rD的不同rnc值的总和

SELECT  t1.*, ca.expectedResult
FROM    myTable t1
CROSS APPLY (
    SELECT SUM(d) expectedResult 
    FROM (SELECT DISTINCT rnc, d 
          FROM myTable t2 
          WHERE t2.rd = t1.rd) AS s) ca

SQL Fiddle

Answer 4

如果你不想表现，很容易

首先你需要一个不同的rD，rnd和d值表
然后你需要将这些值相加
然后将它们加入原来的表格

select rN, rD, rnc, d, sub.dsum as expectedResult
from yourtable
join (
  select rD, rnd, sum(d) as dsum 
  from (select distinct rD, rnd, d
      from yourtable
  ) x
  group by rD, rnd
) sub on yourtable.rD = sub.rD and yourtable.rnd = sub.rnd

SQL SERVER - 对不同

4 个答案: