SQL Server 如何获取不包括周末和节假日的日期差异？答案

【问题标题】：SQL Server How to get Date Difference excluding Weekends and Holidays?SQL Server 如何获取不包括周末和节假日的日期差异？
【发布时间】：2017-04-19 10:02:02
【问题描述】：

我有这张桌子：

tbl_Sales

----------------------------------------
|Item_Code|Sold|Date                   |
|---------+----+-----------------------|
|BBPen100 |30  |2017-04-17 00:00:00.000|
|BBPen100 |21  |2017-04-13 00:00:00.000|
|BBPen100 |13  |2017-04-12 00:00:00.000|
|XSHIRT80 |0   |2017-04-17 00:00:00.000|
|XSHIRT80 |24  |2017-04-14 00:00:00.000|
|XSHIRT80 |9   |2017-04-13 00:00:00.000|
|XSHIRT80 |5   |2017-04-12 00:00:00.000| 
|YBSHADE7 |0   |2017-04-17 00:00:00.000| 
|YBSHADE7 |6   |2017-04-15 00:00:00.000| 
|YBSHADE7 |0   |2017-04-13 00:00:00.000| 
|YBSHADE7 |11  |2017-04-12 00:00:00.000| 
----------------------------------------

如何获得过去 2 个工作日的最后一个非零的 Sold 值？这意味着我需要排除周末和节假日。我有这张包含假期的表格。

tbl_Holiday

-------------------------
|Holiday_Date           |
|-----------------------|
|2017-04-14 00:00:00.000|
|2017-05-01 00:00:00.000|
|2017-10-18 00:00:00.000|
|2017-12-25 00:00:00.000|
-------------------------

比如今天是2017-04-18，输出应该是这样的：

---------------------
|Item_Code|Last_Sold|
|---------+---------|
|BBPen100 |30       |
|XSHIRT80 |9        |
|YBSHADE7 |0        |
---------------------

目标是获取 LAST 2 个工作日内的最后一个 Sold 值，因此计数从 2017 年 4 月 17 日开始。输出分析：

BBPen100-since it has value from last 1 working day (2017-04-17), that value will be retrieved.
XSHIRT80-Zero value from last 1 working day (2017-04-17)
        -2017-04-16 & 2017-04-15 are weekends
        -2017-04-14 is holiday
        -So value from 2017-04-13 will be retrieved.
YBSHADE7-Zero value from last 1 working day (2017-04-17)
        -2017-04-16 & 2017-04-15 are weekends
        -2017-04-14 is holiday
        -2017-04-13 has Zero value
        -2017-04-12 is beyond Last 2 working days
        -So value retrived should be Zero

目前，我有这个查询：

SELECT Item_Code, Sold AS 'Last_Sold'
FROM tbl_Sales
WHERE CONVERT(date, [DATE]) = CASE
                                WHEN CONVERT(date, [DATE]) = CONVERT(date, DATEADD(day, -1, GETDATE())) THEN CONVERT(date, DATEADD(day, -1, GETDATE()))
                                WHEN CONVERT(date, [DATE]) <> CONVERT(date, DATEADD(day, -1, GETDATE())) THEN CONVERT(date, DATEADD(day, -2, GETDATE()))

当然，这不符合要求。

请帮我解决这个问题。

重要提示：考虑周末假期以及如果我在周末或假期运行程序会怎样。

提前谢谢你。

【问题讨论】：

last 2 working days ... 这包括一周中的哪几天？
例如，如果假期是在周末，那么下周一不是工作日，不是吗？
@TriV 无需担心。在假日表中，它已被应用。如果假期是星期日，那么星期一将是假期表中的一个。但如果是星期六，什么都没有改变。这就是要考虑的问题。
请更新您的问题并显示涵盖每个边缘情况的数据。

标签： sql sql-server getdate

【解决方案1】：

你可以试试

样本数据

DECLARE @SampleData as TABLE (Item_Code varchar(10), Sold int, Date datetime)
Insert into @SampleData VALUES
('BBPen100', 30,'2017-04-17 00:00:00.000'),
('BBPen100', 21,'2017-04-13 00:00:00.000'),
('BBPen100', 13,'2017-04-12 00:00:00.000'),
('XSHIRT80', 0 ,'2017-04-17 00:00:00.000'),
('XSHIRT80', 24,'2017-04-14 00:00:00.000'),
('XSHIRT80', 9 ,'2017-04-13 00:00:00.000'),
('XSHIRT80', 5 ,'2017-04-12 00:00:00.000'),
('YBSHADE7', 0 ,'2017-04-17 00:00:00.000'),
('YBSHADE7', 6 ,'2017-04-15 00:00:00.000'),
('YBSHADE7', 0 ,'2017-04-13 00:00:00.000'),
('YBSHADE7', 11,'2017-04-12 00:00:00.000')

DECLARE @TblHoliday AS TABLE
(
   Holiday_Date date
)

INSERT INTO @TblHoliday
VALUES
('2017-04-14 00:00:00.000'),
('2017-05-01 00:00:00.000'),
('2017-10-18 00:00:00.000'),
('2017-12-25 00:00:00.000')

DECLARE @CurrentDate datetime = '2017-04-18 00:00:00'

您可以在@CurrentDate 之前计算@2PreviousWorkingDays

-- 2 Previous Working Day with out Holiday
DECLARE @2PreviousWorkingDay date = CASE
                         WHEN datepart(dw,@CurrentDate) IN (2,3) THEN dateadd(day,-4, @CurrentDate)  -- 2 previous working day before monday
                         WHEN datepart(dw,@CurrentDate) IN (1) THEN dateadd(day,-3, @CurrentDate)   -- 2 previous working day before sunday
                         ELSE dateadd(day,-2, @CurrentDate) -- other day week
                       END
-- with holiday
SELECT @2PreviousWorkingDay = dateadd(day,0 - (SELECT count(1) FROM @TblHoliday th 
                      WHERE  th.Holiday_Date BETWEEN @2PreviousWorkingDay AND @CurrentDate
                      ANd datepart(dw,th.Holiday_Date) NOT IN (7,1) -- calculate only holiday that isn't weekend
                            )                            
                      , @2PreviousWorkingDay

你想要的结果：

;with temps AS
(
SELECT *, row_number() over(PARTITION BY sd.Item_Code ORDER BY sd.[Date] DESC) AS Rn
FROM @SampleData sd
WHERE sd.[Date] >= @2PreviousWorkingDay -- 2 working days
   AND NOT EXISTS (SELECT 1 FROM @TblHoliday th WHERE th.Holiday_Date = Cast(sd.[Date] AS date)) -- not holiday
   AND datepart(dw,sd.[Date]) NOT IN (7,1) -- not weekend
   AND sd.Sold <> 0 -- not zero sold
)
SELECT sd.Item_Code, ISNULL(t.Sold,0) AS Sold FROM
(
   SELECT DISTINCT sd.Item_Code FROM @SampleData sd
) sd
LEFT JOIN temps t ON t.Item_Code = sd.Item_Code AND t.Rn = 1

演示链接：Rextester

【讨论】：

WITH 旁边的分号出现错误。当我删除它时，仍然错误。
之前必须使用WITH 和;。不是;WITH引起的。
知道了！只是分号前缺少右括号。非常感谢你的救命稻草:)

【解决方案2】：

WITH cte1 AS(
    SELECT t.*
    FROM tbl_Sales t
    LEFT JOIN tbl_Holiday hd
        ON t.Date = hd.Holiday_Date
    WHERE hd.Holiday_Date IS NULL AND
          DATENAME(WEEKDAY, t.Date) <> 'Saturday' AND
          DATENAME(WEEKDAY, t.Date) <> 'Sunday'
),
WITH cte2 AS (
    SELECT *,
           ROW_NUMBER() OVER (PARTITION BY Item_Code ORDER BY Date DESC) rn
    FROM cte1
)
SELECT *
FROM cte2
WHERE rn=1

【讨论】：

谢谢你。但是，这不符合所有要求，例如在过去 2 天内不检索零值，如果超过 2 天，则应该为零
您的问题不清楚。这是我能做的最好的了。