Question

我正在使用一种算法，该算法需要根据工作日数进行计算。

样本数据

CREATE TABLE Holidays
(
    StartDate DATETIME,
    EndDate DATETIME
)

INSERT INTO Holidays VALUES ('2019-07-16', '2019-07-17')






DECLARE @CurrentWeekDay INT

SET DATEFIRST 1
// Returns day of the week as int, ex... Monday = 0, Tue = 1....
SET @CurrentWeekDay = DATEPART (WEEKDAY, GETDATE()) - 1

此操作从星期一开始，直到星期五0、1、2、3、4，再减去一周的总天数

现在我创建了2个新变量

DECLARE @TotalHolidays INT, @IsSameWeek BIT

    // If there is a holiday in the current week get the holiday date, else just cast 0
    SET @IsSameWeek = 
        CASE
            WHEN (SELECT DATEDIFF(WEEK,GETDATE(), (SELECT TOP 1 EndDate FROM Holidays))) = -1 THEN 0
        ELSE 
            1
        END 

    SET @TotalHolidays = 
        CASE 
            WHEN @IsSameWeek = 1 THEN (SELECT DATEDIFF(day, StartDate, EndDate ) FROM Holidays)
        ELSE 
            0
        END

因此，基本上使用该简单公式即可计算出一周中剩余的天数

DECLARE @TotalDays = 5,
SET @TotalDays = @TotalDays - @TotalHolidays - @CurrentWeekDay

这一切都很好，但这仅在假期在周末结束时有效。

Datefirst
M T W T F
0 1 2 3 4

如果星期四和星期五是假期，则该算法效果很好，但是如果星期三是假期，那么它将破坏所有内容

基本上我需要它，所以如果星期三是假期

Datefirst就像

M T W T F
0 1   2 3

全部设置完毕，这就是整个代码

CREATE TABLE Holidays
(
    StartDate DATETIME,
    EndDate DATETIME
)

INSERT INTO Holidays VALUES ('2019-07-18', '2019-07-19')

DECLARE @TotalHolidays INT, @IsSameWeek BIT, @CurrentWeekDay INT

SET DATEFIRST 1
SET @CurrentWeekDay = DATEPART (WEEKDAY, GETDATE()) - 1 


    SET @IsSameWeek = 
        CASE
            WHEN (SELECT DATEDIFF(WEEK,GETDATE(), (SELECT TOP 1 EndDate FROM Holidays))) = -1 THEN 0
        ELSE 
            1
        END 

    SET @TotalHolidays = 
        CASE 
            WHEN @IsSameWeek = 1 THEN (SELECT DATEDIFF(day, StartDate, EndDate ) FROM Holidays)
        ELSE 
            0
        END



DECLARE @DaysLeft INT

SET @DaysLeft = 5 - @TotalHolidays - @CurrentWeekDay

这将在返回2时返回0

sql用datefirst忽略星期几

0 个答案: