Question

以下链接中的表格：

enter image description here

Result列是T-SQL查询的结果，其中每条记录通过帐单反映чprodump-modelч从чemployeeч到另一个的移动：amount从employee移出，金额将填入Amount_Out列，反之亦然。

例如：第3行和第4行显示40件产品hp已从emp1移至emp3通过代码111的帐单。此移动会将emp1的此产品总金额降低为60，并将emp3的同一产品的总金额增加为40。

现在，我需要在查询中计算每条记录中的结果，以显示员工保管产品的总金额，直至记录中的日期为止：

结果=为所有先前账单计算的金额+ Amount_In - Amount_Out

这意味着结果将取决于先前记录的值。如何通过SQL Server 2008中的查询实现这一目标？什么是最好的实现它，存储过程或视图？

Answer 1

您可以使用相关子查询执行此操作。如果AND it.No <= ot.No不是连续的，则将AND it.Date <= ot.Date更改为No：

DECLARE @t TABLE
    (
      No INT ,
      Employee CHAR(4) ,
      Model CHAR(4) ,
      Date DATE ,
      Bill CHAR(3) ,
      Income MONEY ,
      Outcome MONEY
    )

INSERT  INTO @t
VALUES  ( 1, 'emp1', 'hp', '20140101', '000', 100, 0 ),
        ( 2, 'emp1', 'dell', '20140101', '000', 100, 0 ),
        ( 3, 'emp1', 'hp', '20140308', '111', 0, 40 ),
        ( 4, 'emp3', 'hp', '20140308', '111', 40, 0 ),
        ( 5, 'emp3', 'hp', '20140408', '222', 0, 20 ),
        ( 6, 'emp2', 'hp', '20140408', '222', 20, 0 ),
        ( 7, 'emp1', 'hp', '20140608', '333', 0, 5 ),
        ( 8, 'emp2', 'hp', '20140608', '333', 5, 0 ),
        ( 9, 'emp1', 'dell', '20150101', '444', 0, 40 )


SELECT  * ,
        ( SELECT    SUM(Income - Outcome)
          FROM      @t it
          WHERE     it.Model = ot.Model
                    AND it.Employee = ot.Employee
                    AND it.No <= ot.No
        ) AS Result
FROM    @t ot

输出：

No  Employee    Model   Date        Bill    Income  Outcome Result
1   emp1        hp      2014-01-01  000     100.00  0.00    100.00
2   emp1        dell    2014-01-01  000     100.00  0.00    100.00
3   emp1        hp      2014-03-08  111     0.00    40.00   60.00
4   emp3        hp      2014-03-08  111     40.00   0.00    40.00
5   emp3        hp      2014-04-08  222     0.00    20.00   20.00
6   emp2        hp      2014-04-08  222     20.00   0.00    20.00
7   emp1        hp      2014-06-08  333     0.00    5.00    55.00
8   emp2        hp      2014-06-08  333     5.00    0.00    25.00
9   emp1        dell    2015-01-01  444     0.00    40.00   60.00

Answer 2

Sql 2008无法访问Sql2012中的其他分析函数，例如LEAD和LAG，因此您需要采用these here之一的解决方法

这是递归CTE实现的一个例子。

请注意，这取决于能否对Employee，ProductModel分组中的连续移动进行逐步编号（1,2,3等）而没有任何歧义，因为这用于JOIN条件（Sequence to Sequence + 1）。例如，如果存在平局，您将获得不稳定的结果。

WITH RankedByEmpProduct AS
(
    SELECT No, Employee, Product_Model, Date, Bill_Code, Amount_In, 
           Amount_Out, Amount_In-Amount_Out AS Nett,
           DENSE_RANK() OVER (PARTITION BY Employee, Product_Model 
           ORDER BY No ASC) AS EmpModelSeq
    FROM ProductModel
),
RecursiveRunningTotal AS 
(
    SELECT No, Employee, Product_Model, Date, Bill_Code, Amount_In, 
           Amount_Out, EmpModelSeq, Amount_In-Amount_Out AS Nett, 
           Amount_In-Amount_Out AS RunningTotal
    FROM RankedByEmpProduct rep
    WHERE EmpModelSeq = 1

    UNION ALL

    SELECT rep.No, rep.Employee, rep.Product_Model, rep.Date, rep.Bill_Code, 
           rep.Amount_In, rep.Amount_Out, rep.EmpModelSeq, 
           rep.Amount_In-rep.Amount_Out AS Nett, 
           rrt.RunningTotal + rep.Amount_In - rep.Amount_Out AS RunningTotal
    FROM RankedByEmpProduct rep
        INNER JOIN RecursiveRunningTotal rrt
        ON rep.Employee = rrt.Employee AND rep.Product_Model = rrt.Product_Model 
        AND rep.EmpModelSeq = rrt.EmpModelSeq + 1
)
SELECT * FROM RecursiveRunningTotal
ORDER BY No ASC;

SqlFiddle here

Answer 3

您可以通过简单的子查询获取结果。

SELECT No,Employee,Product_Model,[Date],Bill_Code,Amount_In,Amount_Out,
(
    SELECT SUM
    (
        SELECT D1.Amount_In - D0.Amount_Out AS subTotal
        FROM Table_Name AS D1
        WHERE 
        (
            D1.[Date] < D0.[Date]
        )
        OR
        (
            D1.[Date] = D0.[Date]
            AND 
            D1.No <= D0.No
        )
    )
) AS Result
FROM Table_Name D0
ORDER BY D0.No, D0.[Date]

实际输出应为：

No  Employee    Product_Model   Date        Bill_Code    Amount_In  Amount_Out   Result
1   emp1        hp              2014-01-01  000          100.00     0.00         100.00
2   emp1        dell            2014-01-01  000          100.00     0.00         200.00
3   emp1        hp              2014-03-08  111          0.00       40.00        160.00
4   emp3        hp              2014-03-08  111          40.00      0.00         200.00
5   emp3        hp              2014-04-08  222          0.00       20.00        180.00
6   emp2        hp              2014-04-08  222          20.00      0.00         200.00
7   emp1        hp              2014-06-08  333          0.00       5.00         195.00
8   emp2        hp              2014-06-08  333          5.00       0.00         200.00
9   emp1        dell            2015-01-01  444          0.00       40.00        160.00

根据以前的行计算列

3 个答案: