Question

这是我创建的一个表，用于解释我想要做的事情：

create table #test (
    PlaceID int, 
    ItemID int, 
    ItemCount int, 
    Amount dec(11,2)
)

我想得到3件事：

sum by Place
按地点和项目总和
按地点和非项目总和

前两个很简单：

sum(Amount) over (partition by PlaceID) as PlaceAmount
sum(Amount) over (partition by PlaceID, ItemID) as PlaceItemAmount

但是，如何获得当前项目中不是当前项目的所有项目的总和？

以下是设置了数据和查询的SQL Fiddle：

Answer 1

select t1.PlaceID, t1.ItemID, t1.ItemCount
     , t1.Amount as 'AmtMe'
     , SumPlace.sum as 'AmtPlace' 
     , SumPlace.sum - t1.Amount as 'AmtPlaceNoMe'
  from #test as t1
  join (select PlaceID, sum(Amount) as 'sum'
          from #test 
         group by PlaceID) as SumPlace 
    on t1.PlaceID = SumPlace.PlaceID

Answer 2

这是否符合您的期望？基本上，取整个（按地点分区）并减去当前（按地点，项目分区）以获得余数。但是，我会提到将它保存在子查询中，以便每个函数和分区集只运行一次窗口聚合。

同样可以使用该逻辑进行计数。

select
    PlaceID,
    ItemID,
    ItemCount,
    Amount,
    PlaceItemCount,
    PlaceAmount,
    ItemAndPlaceAmount,
    PlaceAmount-ItemAndPlaceAmount as RemainderAmount
from (
  select
      PlaceID,
      ItemID,
      ItemCount,
      Amount,
      sum(ItemCount) over (partition by PlaceID) as PlaceItemCount,
      sum(Amount) over (partition by PlaceID) as PlaceAmount,
      sum(Amount) over (partition by PlaceID, ItemID) as ItemAndPlaceAmount
  from tblTest
  ) z

Answer 3

SELECT
    PlaceID,
    ItemID,
    ItemCount,
    Amount,
    sum(ItemCount) over (partition BY PlaceID) AS PlaceItemCount,
    sum(Amount) over (partition BY PlaceID) AS PlaceAmount
  , sum(Amount) over (partition BY PlaceID, ItemID) AS PlaceItemAmount
  , sum(Amount) over (partition BY PlaceID)
  - sum(Amount) over (partition BY PlaceID, ItemID) AS PlaceItemAmountMinusGroup
  , sum(Amount) over (partition BY PlaceID) - Amount PlaceItemAmountMinusThis
FROM tblTest

PlaceItemAmountMinusGroup是按地点计算的总金额，不包括ItemID的总金额 PlaceItemAmountMinusThis是没有行数的地方总金额。

SQLFiddle demo

是否可以在窗口函数分区中进行过滤

3 个答案: