Question

创建表格假日主人，其名称为假日日期，日期（星期日，星期一或任何公众假期）。

创建一个将返回4个输出列的函数：Date，day of month，year＆amp;天。日期将是2个日期之间的工作日期（忽略假日主表中的日期）

CREATE function DATENEWww
AS 
BEGIN
DECLARE @StartDate DATETIME
DECLARE @MIDDLEDATE DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = '2013/05/01'
SET @EndDate = '2013/05/10'

SELECT getday()'Day', getDATE() 'Today', YEAR(GETDATE()) 'YEAR Part'
SELECT 
   (DATEDIFF(dd, @StartDate, @EndDate) + 1)
  -(DATEDIFF(wk, @StartDate, @EndDate) * 2)
  -(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN 1 ELSE 0 END)
  -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END)
  -(CASE WHEN DATENAME (DW, @MIDDLEDATE)='SELECT HOLIDAY_DATE FROM HOLIDAY_MASTER' THEN 1 ELSE 0 END)

  END

Answer 1

使用表格功能：

CREATE FUNCTION [dbo].[CalculateNumberOFWorkDays] (@StartDate datetime, @EndDate datetime)
RETURNS RETURNS @Results TABLE (
    [Date]         datetime,
    [DayOfMonth]   int,
    [Year]         int,
    [Day]          int
)
AS BEGIN

SET @StartDate = DATEADD(dd, DATEDIFF(dd, 0, @StartDate), 0) 
SET @EndDate = DATEADD(dd, DATEDIFF(dd, 0, @EndDate), 0) 

DECLARE @WORKDAYS INT
SELECT @WORKDAYS = (DATEDIFF(dd, @StartDate, @EndDate) + 1)
           -(DATEDIFF(wk, @StartDate, @EndDate) * 2)
           -(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN 1 ELSE 0 END)
           -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END)


INSERT @Results ([Date], [DayOfMonth], [Year], [Day])
SELECT <your query here>

RETURN 
END

创建一个将返回4个输出列的函数

1 个答案: