SQL Server查询以获取日期范围内所有月份的第二个最后一个工作日的列表，其中包括日历中的假期

Question

我需要编写一个查询来获取：

1. 日期结果集。结果集包含从当前日期前 3 个月开始的所有日期 + 当前日期后 12 个月（包括当前月份）的所有日期。例如：说当前日期：7/25/2018 结果集应包含从 2018 年 4 月 1 日到 2019 年 6 月 30 日的所有日期

2. 从此结果集中，将获得每个月的倒数第二个工作日。

条件：倒数第二个工作日应为平日和非节假日。假期列表包含在名为RT_STATEHOLIDAY_MV 的表中。

RT_STATEHOLIDAY_MV：

Description
-----------
01/02/17
01/16/17
02/28/17
04/27/18
05/29/17
07/04/17
09/04/17
11/10/17
11/23/17
12/25/17

到目前为止我所尝试的：

WITH cte AS 
(
    SELECT 
        1 AS DayID,
        DATEADD(month, -3, DATEADD(month, DATEDIFF(month, 0, GETDATE()), 0)) AS FromDate,
        DATENAME(dw, DATEADD(month, -3, DATEADD(month, DATEDIFF(month, 0, GETDATE()), 0))) AS Dayname
    UNION ALL
    SELECT 
        cte.DayID + 1 AS DayID,
        DATEADD(d, 1, cte.FromDate),
        DATENAME(dw, DATEADD(d, 1, cte.FromDate)) AS Dayname
    FROM 
        cte
    WHERE 
        DATEADD(d, 1, cte.FromDate) < DATEADD(month, 12, DATEADD(month, DATEDIFF(month, 0, GETDATE()), 0))
) ,
wrkInd as 
(
    SELECT 
        (SELECT CONVERT(VARCHAR(10), CTE.FromDate, 1) AS [MMDDYY]) AS Date, 
        CASE 
           WHEN REFSH.DESCRIPTION IS NULL THEN 'Y' ELSE 'N' END AS WORKDAY_INDICATOR
    FROM 
        CTE
    LEFT JOIN 
        RT_STATEHOLIDAY_MV REFSH ON CTE.FromDate =  REFSH.DESCRIPTION
    WHERE 
        DayName NOT IN ('Saturday','Sunday')
),
secLastWrkDay AS
(
    SELECT 
        *,
        CASE WHEN wrkInd.Date = 
--(SELECT MAX(wrkInd.Date) 
--from wrkInd where wrkInd.Date <=
(select case 
    datename(WEEKDAY,EOMONTH(wrkInd.Date)) 
    when 'Saturday' then convert(datetime,(EOMONTH(wrkInd.Date)))-2
    when 'Sunday' then  convert(datetime, (EOMONTH(wrkInd.Date)))-3 
    when 'Monday' then convert(datetime, (EOMONTH(wrkInd.Date)))-3 
    else  convert(datetime,EOMONTH(wrkInd.Date))-1 
end) 
-- AND wrkInd.WORKDAY_INDICATOR='Y'
-- AND MONTH(tab)=MONTH(wrkInd.Date)
--)
THEN 'Y' ELSE 'N' END AS SECOND_LAST_WRK_DAY
FROM wrkInd
),
cutoff AS
(
    SELECT 
        secLastWrkDay.Date, secLastWrkDay.WORKDAY_INDICATOR,
        CASE 
           WHEN MONTH(secLastWrkDay.Date) = MONTH(EOMONTH(secLastWrkDay.Date))
                AND YEAR(secLastWrkDay.Date) = YEAR(EOMONTH(secLastWrkDay.Date))
                AND secLastWrkDay.WORKDAY_INDICATOR = 'Y'
                AND secLastWrkDay.SECOND_LAST_WRK_DAY = 'Y'
               THEN 'Y' 
               ELSE 'N' 
        END AS SECOND_LAST_WORK_DAY
    FROM 
        secLastWrkDay
)
SELECT * 
FROM cutoff
OPTION (MaxRecursion 700);

这可行，但这并没有考虑假期表中的假期。

Answer 1

问题是您要根据工作日减去 1、2 或 3 天，而不考虑节假日。 顺便说一句，您正在过滤掉周末，因此 secLastWrkDay 中的“周六”和“周日”案例将永远不会发生。 您可以使用每年/每月（和 WORKDAY_INDICATOR）剩余天数的反向编号，并选择第 2 个工作日作为您想要的工作日。另外，我建议不要将 varchar 用于日期值。您的查询可能如下所示：

WITH 
  cte (FromDate) AS (
    SELECT DATEADD(month, DATEDIFF(month, 0, GETDATE()) - 3, 0)
    UNION ALL
    SELECT DATEADD(d, 1, cte.FromDate)
    FROM cte
    WHERE DATEADD(d, 1, cte.FromDate) < DATEADD(month, DATEDIFF(month, 0, GETDATE()) + 12, 0)
  ),
  wrkInd ([Date], WORKDAY_INDICATOR) AS (
    SELECT cte.FromDate, 
      CASE WHEN REFSH.DESCRIPTION IS NULL THEN 'Y' ELSE 'N' END
    FROM CTE
      LEFT JOIN RT_STATEHOLIDAY_MV REFSH ON CTE.FromDate =  REFSH.DESCRIPTION
    WHERE DATENAME(dw, cte.FromDate) NOT IN ('Saturday','Sunday')
  ),
  dayNumber ([Date], WORKDAY_INDICATOR, dayNumber) AS (
    SELECT [Date], WORKDAY_INDICATOR
      , ROW_NUMBER() OVER 
        (PARTITION BY YEAR([Date]), MONTH([Date]), WORKDAY_INDICATOR ORDER BY [Date] DESC)
    FROM wrkInd
  ),
  cutoff ([Date], WORKDAY_INDICATOR, SECOND_LAST_WORK_DAY) AS (
    SELECT [Date], WORKDAY_INDICATOR
      , CASE 
          WHEN WORKDAY_INDICATOR = 'Y' AND dayNumber = 2
          THEN 'Y' ELSE 'N' 
        END
    FROM dayNumber
  )
SELECT * FROM cutoff ORDER BY [Date]
OPTION (MaxRecursion 700);

Answer 2

看看这个。如果我能正确理解你

WITH CTE AS
(
SELECT Cast(DATEADD(month, -3, GETDATE()) AS DATE) AS D
UNION ALL 
SELECT
DATEADD(day, 1, C.D) AS D
FROM
CTE C
WHERE
C.D<Cast(DATEADD(month, 11, EOMONTH(GETDATE())) AS DATE)  
)
SELECT
C.D
FROM
CTE C
LEFT JOIN RT_STATEHOLIDAY_MV H ON H.D=C.D
WHERE
datename(WEEKDAY,C.D) NOT IN ('Saturday', 'Sunday')
AND H.D IS NULL
OPTION (MaxRecursion 1000);

关于小提琴http://sqlfiddle.com/#!18/52aff/24/0的示例