Question

我正在编写一个具有2个参数的函数，即datenow和'3'，用于添加日期+ 3个工作日。

我已经成功地为周末添加了date + 3个工作日，但是对于假期，结果却不是我期望的，对于假期，我有一个名为lkp_holiday的表想法是获取当前日期并检查lkp_holiday上的日期，如果有相同的日期，则添加addDate

ALTER FUNCTION [dbo].[DATEADDEXCLUDEWD](@addDate AS DATE, @numDays AS INT)
RETURNS DATETIME
AS
BEGIN
  DECLARE @DateHoliday DATETIME

  WHILE @numDays > 0
  BEGIN
    SET @addDate = DATEADD(d, 1, @addDate)
      --For weekend
    IF DATENAME(DW, @addDate) = 'saturday' SET @addDate = DATEADD(d, 1, @addDate)
    IF DATENAME(DW, @addDate) = 'sunday' SET @addDate = DATEADD(d, 1, @addDate)
      --For Holiday
      IF EXISTS(SELECT DISTINCT hol_date  FROM [Vacation].[dbo].[Lkp_Holiday] WHERE hol_date > GETDATE())
      BEGIN
          DECLARE M_CURSOR CURSOR
          FOR SELECT DISTINCT hol_date  FROM [Vacation].[dbo].[Lkp_Holiday] WHERE hol_date > GETDATE()

          OPEN M_CURSOR
          FETCH NEXT FROM M_CURSOR INTO @DateHoliday

          WHILE @@FETCH_STATUS = 0
          BEGIN
              SET @addDate = DATEADD(DAY, 1, @addDate)

              FETCH NEXT FROM M_CURSOR INTO @DateHoliday
          END
          CLOSE M_CURSOR
          DEALLOCATE M_CURSOR
      END                                         

    SET @numDays = @numDays - 1
  END

  RETURN CAST(@addDate AS DATETIME)
END

例如今天是2019-05-17，加上3天，输出2019-05-22 = >>是真的，因为它包括周末

我已经运行了该函数，并且在lkp_holiday表中我有一个约会假期，比如说2019-05-23

我期望的是2019-05-24，但是此函数的结果是2019-05-25

Answer 1

正如其他人指出的那样，这远不是解决此问题的最佳方法。但是，随着代码的正常运行，存在以下问题，在以下代码中已得到纠正：

您需要检查假日日期是否是循环所处理的日期，否则，您需要在每次循环中添加假日日期。
为此，您需要将@DateHoliday变量的类型设置为date而不是datetime。
为确保周末跳过代码有效，您需要在添加
“假期”在“ 2019-05-23”的“ 2019-05-17”输出仍然是“ 2019-05-22”，但是现在“ 2019-05-18”的输出“ 2019-05-24”，即添加了另一天作为假日。

    ALTER FUNCTION [dbo].[DATEADDEXCLUDEWD]
    (
      @addDate AS DATE
      , @numDays AS INT
    )
    RETURNS DATETIME
    AS
    BEGIN
      -- Needs to be a date type to allow for a date to date compare in the holiday section
      DECLARE @DateHoliday DATE

      WHILE @numDays > 0
      BEGIN
          --For weekend

        -- Add these before the regular add date, as otherwise we've already moved the date forward 1 day
        IF DATENAME(DW, @addDate) = 'saturday' SET @addDate = DATEADD(d, 1, @addDate)
        IF DATENAME(DW, @addDate) = 'sunday' SET @addDate = DATEADD(d, 1, @addDate)

        SET @addDate = DATEADD(d, 1, @addDate)

          --For Holiday
          IF EXISTS(SELECT DISTINCT hol_date FROM [Vacation].[dbo].[Lkp_Holiday] WHERE hol_date > GETDATE())
          BEGIN
              DECLARE M_CURSOR CURSOR
              FOR SELECT DISTINCT hol_date FROM [Vacation].[dbo].[Lkp_Holiday] WHERE hol_date > GETDATE()

              OPEN M_CURSOR
              FETCH NEXT FROM M_CURSOR INTO @DateHoliday

              WHILE @@FETCH_STATUS = 0
              BEGIN
            -- Only add the day if we've on the holiday day
            if @DateHoliday = @addDate begin
                  SET @addDate = DATEADD(DAY, 1, @addDate)
            end

                  FETCH NEXT FROM M_CURSOR INTO @DateHoliday
              END
              CLOSE M_CURSOR
              DEALLOCATE M_CURSOR
          END                                     

        SET @numDays = @numDays - 1
      END

      RETURN CAST(@addDate AS DATETIME)
    END

用于添加日期+3个工作日（不包括周末和节假日）的功能

1 个答案: