Question

如何让下表仅显示具有最早日期的行以及新列中具有相同项目ID的“数量”总和。

如下图所示，

Name  ItemID   Date         Qty
A     A0001    25-Feb-14    23
B     A0001    01-Apr-14    4
C     A0001    18-Jul-14    54
D     K0006    03-Mar-14    86
E     K0006    11-Jan-14    29
F     X0003    27-Sep-14    19
G     X0003    12-Nov-14    28

这就是结果应该是什么样的，

Name   ItemID   Date        Qty  Total Outstanding
A      A0001    25-Feb-14   23   81
E      K0006    11-Jan-14   29   115
F      X0003    27-Sep-14   19   47

Answer 1

SELECT a.Name,a.ItemID,b.Date,a.Qty,b.Total FROM TableA a
INNER JOIN (SELECT ItemID,MIN(Date) AS Date,SUM(Qty)
AS Tatal FROM TableA GROUP BY ItemID)b ON
b.ItemID=a.ItemID AND a.Date=b.Date

Answer 2

DECLARE @T TABLE (NAME VARCHAR(10),Itemid VARCHAR(10),da date,qty int)

insert into @T(NAME,Itemid,da,qty) values ('A','A0001','2014-02-25',23),
('B','A0001','2014-04-01',4),
('C','A0001','2014-07-18',54),
('D','K0006','2014-03-03',86),
('E','K0006','2014-01-11',29),
('F','X0003','2014-09-27',19),
('G','X0003','2014-11-11',28)




 ;with cte as (
    select *,SUM(qty)OVER(PARTITION BY Itemid  )total,
    ROW_NUMBER()OVER(PARTITION BY Itemid ORDER BY da )RN from @T
    group by NAME,Itemid,da,qty
    )

    select NAME,Itemid,convert(varchar,da,100)As [date],qty,total from cte 
    WHERE RN = 1

Answer 3

试试这个：

with CTE
as(
select  -- M.Name,
SUM(M.Qty) TotalOTD,M.ItemID,MIN(M.Date) as MDate from myTestTable M
GROUP BY M.ItemID )
select Name,    D.ItemID,  Date ,   Qty, TotalOTD from CTE
JOIN myTestTable D
ON CTE.ItemID=D.ItemID and CTE.MDate=D.date

输出

Name    ItemID  Date        Qty TotalOTD
A      A0001    2014-02-25  23  81
E      K0006    2014-01-11  29  115
F      X0003    2014-09-27  19  47

Answer 4

Select Name, AliasID, DateAlias, Quantity, TotalOutstanding 
from TableA A join 
        (Select ItemId as AliasID,Min(Date) as DateAlias,
        Sum(Quantity) as TotalOutstanding 
        from TableA
        group by ItemID )V
        on A.ItemID=V.ItemID

对SQL表进行排序以显示第一行和所有其他行的总和

4 个答案: