Question

我正在使用SQL Server 2008，我遇到了这个问题。

我有一个包含工作字段，日期字段和成本字段的表。日期字段包含在特定作业上完成工作的日期，而成本字段包含在该日期花费的金额（单个作业可以在几天内完成工作）

我想创建一个select语句，以便获得在不同列中打开作业的0-7days和8-30days之间花费的总和。

我试过了：

select a.[Job No_], DATEDIFF(day, MIN(b.[Posting Date]), a.[Ending Date]) as [WIP Age],
       SUM(b.[Total Cost])[Total WIP cost]
  [0 - 7 days]=case
         when (DATEDIFF(day, MIN(b.[Posting Date]), a.[Ending Date]) between 1 and 7) then
            SUM(b.[Total Cost])
        else 0
    end,
    [8 - 30 days]=case
        when (DATEDIFF(day, MIN(b.[Posting Date]), a.[Ending Date]) between 8 and 30) then
            SUM(b.[Total Cost])
        else 0
    end,
    [> 30 days]=case
        when (DATEDIFF(day, MIN(b.[Posting Date]), a.[Ending Date]) > 30) then
            SUM(b.[Total Cost])
        else 0
    end
 from [Job]a 
 right join [Job Ledger Entry]b
 on b.[Job No_] = a.[No_] 
 group by a.[No_]

显示与此类似的结果：

Job No_  WIP Age    Total WIP cost    0 - 7 days   8 - 30 days   > 30 days
 0001      40           9526.18          0           0             9526.18
 0002      27           1178.49          0        1178.49            0
 0003      3            2838.94       2838.94        0               0

这不是我想要的，我希望所有费用分别在0 - 7天，8 - 30天和> 30天之间。

Answer 1

以下可能会做得更好：

SELECT
  a.[Job No_],
  [WIP Age] = DATEDIFF(day, MIN(b.[Posting Date]), MAX(a.[Ending Date])),
  [Total WIP cost] = SUM(b.[Total Cost]),
  [0 - 7 days] = SUM(CASE
    WHEN (DATEDIFF(DAY, b.[Posting Date], a.[Ending Date]) BETWEEN 1 AND 7)
    THEN SUM(b.[Total Cost])
  END),
  [8 - 30 days] = SUM(CASE
    WHEN (DATEDIFF(DAY, b.[Posting Date], a.[Ending Date]) BETWEEN 8 AND 30)
    THEN b.[Total Cost]
  END),
  [> 30 days] = SUM(CASE
    WHEN (DATEDIFF(day, b.[Posting Date], a.[Ending Date]) > 30)
    THEN b.[Total Cost]
  END)
FROM [Job] a
  RIGHT JOIN [Job Ledger Entry] b ON b.[Job No_] = a.[No_]
GROUP BY a.[No_]

说明：我将所有CASE表达式放在相应的SUM中。我还在第一个DATEDIFF中将MAX（）应用于a.[Ending Date]，因为使用该列而不进行聚合似乎是错误的。

如何获得不同时期之间的金额总和

1 个答案: