计算两个日期之间的工作日答案

【问题标题】：Count work days between two dates计算两个日期之间的工作日
【发布时间】：2010-09-20 03:10:55
【问题描述】：

如何计算 SQL Server 中两个日期之间的工作日数？

周一到周五，必须是 T-SQL。

【问题讨论】：

你能定义工作日吗？周一到周五有吗？不包括重大节日？什么国家？必须在 SQL 中完成吗？

【解决方案1】：

以上功能均不适用于同一周或处理假期。我是这样写的：

create FUNCTION [dbo].[ShiftHolidayToWorkday](@date date)
RETURNS date
AS
BEGIN
    IF DATENAME( dw, @Date ) = 'Saturday'
        SET @Date = DATEADD(day, - 1, @Date)

    ELSE IF DATENAME( dw, @Date ) = 'Sunday'
        SET @Date = DATEADD(day, 1, @Date)

    RETURN @date
END
GO

create FUNCTION [dbo].[GetHoliday](@date date)
RETURNS varchar(50)
AS
BEGIN
    declare @s varchar(50)

    SELECT @s = CASE
        WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year]  ) + '-01-01') = @date THEN 'New Year'
        WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year]+1) + '-01-01') = @date THEN 'New Year'
        WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year]  ) + '-07-04') = @date THEN 'Independence Day'
        WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year]  ) + '-12-25') = @date THEN 'Christmas Day'
        --WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year]) + '-12-31') = @date THEN 'New Years Eve'
        --WHEN dbo.ShiftHolidayToWorkday(CONVERT(varchar, [Year]) + '-11-11') = @date THEN 'Veteran''s Day'

        WHEN [Month] = 1  AND [DayOfMonth] BETWEEN 15 AND 21 AND [DayName] = 'Monday' THEN 'Martin Luther King Day'
        WHEN [Month] = 5  AND [DayOfMonth] >= 25             AND [DayName] = 'Monday' THEN 'Memorial Day'
        WHEN [Month] = 9  AND [DayOfMonth] <= 7              AND [DayName] = 'Monday' THEN 'Labor Day'
        WHEN [Month] = 11 AND [DayOfMonth] BETWEEN 22 AND 28 AND [DayName] = 'Thursday' THEN 'Thanksgiving Day'
        WHEN [Month] = 11 AND [DayOfMonth] BETWEEN 23 AND 29 AND [DayName] = 'Friday' THEN 'Day After Thanksgiving'
        ELSE NULL END
    FROM (
        SELECT
            [Year] = YEAR(@date),
            [Month] = MONTH(@date),
            [DayOfMonth] = DAY(@date),
            [DayName]   = DATENAME(weekday,@date)
    ) c

    RETURN @s
END
GO

create FUNCTION [dbo].GetHolidays(@year int)
RETURNS TABLE 
AS
RETURN (  
    select dt, dbo.GetHoliday(dt) as Holiday
    from (
        select dateadd(day, number, convert(varchar,@year) + '-01-01') dt
        from master..spt_values 
        where type='p' 
        ) d
    where year(dt) = @year and dbo.GetHoliday(dt) is not null
)

create proc UpdateHolidaysTable
as

if not exists(select TABLE_NAME from INFORMATION_SCHEMA.TABLES where TABLE_NAME = 'Holidays')
    create table Holidays(dt date primary key clustered, Holiday varchar(50))

declare @year int
set @year = 1990

while @year < year(GetDate()) + 20
begin
    insert into Holidays(dt, Holiday)
    select a.dt, a.Holiday
    from dbo.GetHolidays(@year) a
        left join Holidays b on b.dt = a.dt
    where b.dt is null

    set @year = @year + 1
end

create FUNCTION [dbo].[GetWorkDays](@StartDate DATE = NULL, @EndDate DATE = NULL)
RETURNS INT 
AS
BEGIN
    IF @StartDate IS NULL OR @EndDate IS NULL
        RETURN  0

    IF @StartDate >= @EndDate 
        RETURN  0

    DECLARE @Days int
    SET @Days = 0

    IF year(@StartDate) * 100 + datepart(week, @StartDate) = year(@EndDate) * 100 + datepart(week, @EndDate) 
        --same week
        select @Days = (DATEDIFF(dd, @StartDate, @EndDate))
      - (CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN 1 ELSE 0 END)
      - (CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END)
      - (select count(*) from Holidays where dt between @StartDate and @EndDate)
    ELSE
        --diff weeks
        select @Days = (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)
      - (select count(*) from Holidays where dt between @StartDate and @EndDate)
 
    RETURN  @Days
END

【讨论】：

我刚刚在代码项目codeproject.com/Tips/5284659/…的一篇文章中添加了这段代码

【解决方案2】：

我借鉴了其他人的一些想法来创建我的解决方案。我使用内联代码来忽略周末和美国联邦假期。在我的环境中，EndDate 可能为空，但它永远不会在 StartDate 之前。

CREATE FUNCTION dbo.ufn_CalculateBusinessDays(
@StartDate DATE,
@EndDate DATE = NULL)

RETURNS INT
AS

BEGIN
DECLARE @TotalBusinessDays INT = 0;
DECLARE @TestDate DATE = @StartDate;


IF @EndDate IS NULL
    RETURN NULL;

WHILE @TestDate < @EndDate
BEGIN
    DECLARE @Month INT = DATEPART(MM, @TestDate);
    DECLARE @Day INT = DATEPART(DD, @TestDate);
    DECLARE @DayOfWeek INT = DATEPART(WEEKDAY, @TestDate) - 1; --Monday = 1, Tuesday = 2, etc.
    DECLARE @DayOccurrence INT = (@Day - 1) / 7 + 1; --Nth day of month (3rd Monday, for example)

    --Increment business day counter if not a weekend or holiday
    SELECT @TotalBusinessDays += (
        SELECT CASE
            --Saturday OR Sunday
            WHEN @DayOfWeek IN (6,7) THEN 0
            --New Year's Day
            WHEN @Month = 1 AND @Day = 1 THEN 0
            --MLK Jr. Day
            WHEN @Month = 1 AND @DayOfWeek = 1 AND @DayOccurrence = 3 THEN 0
            --G. Washington's Birthday
            WHEN @Month = 2 AND @DayOfWeek = 1 AND @DayOccurrence = 3 THEN 0
            --Memorial Day
            WHEN @Month = 5 AND @DayOfWeek = 1 AND @Day BETWEEN 25 AND 31 THEN 0
            --Independence Day
            WHEN @Month = 7 AND @Day = 4 THEN 0
            --Labor Day
            WHEN @Month = 9 AND @DayOfWeek = 1 AND @DayOccurrence = 1 THEN 0
            --Columbus Day
            WHEN @Month = 10 AND @DayOfWeek = 1 AND @DayOccurrence = 2 THEN 0
            --Veterans Day
            WHEN @Month = 11 AND @Day = 11 THEN 0
            --Thanksgiving
            WHEN @Month = 11 AND @DayOfWeek = 4 AND @DayOccurrence = 4 THEN 0
            --Christmas
            WHEN @Month = 12 AND @Day = 25 THEN 0
            ELSE 1
            END AS Result);

    SET @TestDate = DATEADD(dd, 1, @TestDate);
END

RETURN @TotalBusinessDays;
END

【讨论】：

【解决方案3】：

这基本上是 CMS 的答案，不依赖于特定的语言设置。由于我们正在拍摄通用，这意味着它也应该适用于所有 @@datefirst 设置。

datediff(day, <start>, <end>) + 1 - datediff(week, <start>, <end>) * 2
    /* if start is a Sunday, adjust by -1 */
  + case when datepart(weekday, <start>) = 8 - @@datefirst then -1 else 0 end
    /* if end is a Saturday, adjust by -1 */
  + case when datepart(weekday, <end>) = (13 - @@datefirst) % 7 + 1 then -1 else 0 end

datediff(week, ...) 在几周内始终使用周六到周日的边界，因此该表达式是确定性的，不需要修改（只要我们对工作日的定义始终是周一到周五）。日期编号确实有所不同根据@@datefirst 设置和修改后的计算处理此校正，并带有一些模运算的小复杂性。

处理周六/周日事情的一种更简洁的方法是在提取星期几值之前翻译日期。移位后，这些值将重新与固定的（可能更熟悉的）编号保持一致，该编号在周日以 1 开始，在周六以 7 结束。

datediff(day, <start>, <end>) + 1 - datediff(week, <start>, <end>) * 2
  + case when datepart(weekday, dateadd(day, @@datefirst, <start>)) = 1 then -1 else 0 end
  + case when datepart(weekday, dateadd(day, @@datefirst, <end>))   = 7 then -1 else 0 end

至少早在 2002 年和 Itzik Ben-Gan 的一篇文章中，我就已经跟踪过这种形式的解决方案。 (https://technet.microsoft.com/en-us/library/aa175781(v=sql.80).aspx) 虽然由于较新的 date 类型不允许日期算术，但它需要稍作调整，但在其他方面是相同的。

编辑：我添加了不知何故被遗漏的+1。还值得注意的是，此方法始终计算开始天数和结束天数。它还假定结束日期在开始日期之前或之后。

【讨论】：

请注意，这将在周末的许多日期返回错误的结果，因此它们不会加起来（周五->周一应该与周五->周六+周六->周日+周日->周一相同）。 Fri->Sat 应为 0（正确），Sat->Sun 应为 0（错误 -1），Sun->Mon 应为 1（错误 0）。其他错误是 Sat->Sat = -1, Sun->Sun = -1, Sun->Sat = 4
@adrianm 我相信我已经纠正了这些问题。实际上问题是它总是被一个关闭，因为我不小心把那个部分掉了。
感谢您的更新。我以为你的公式不包括我需要的开始日期。自己解决并添加为另一个答案。

【解决方案4】：

我知道这是一个老问题，但我需要一个不包括开始日期的工作日公式，因为我有几个项目并且需要正确累积天数。

没有一个非迭代的答案对我有用。

我使用了类似的定义

午夜到周一、周二、周三、周四和周五的次数

（其他人可能会从午夜计算到星期六而不是星期一）

我最终得到了这个公式

SELECT DATEDIFF(day, @StartDate, @EndDate) /* all midnights passed */
     - DATEDIFF(week, @StartDate, @EndDate) /* remove sunday midnights */
     - DATEDIFF(week, DATEADD(day, 1, @StartDate), DATEADD(day, 1, @EndDate)) /* remove saturday midnights */

【讨论】：

那个是为我做的，但我必须做一点小改动。它没有考虑@StartDate 是星期六还是星期五。这是我的版本：DATEDIFF(day, @StartDate, @EndDate) - DATEDIFF(week, @StartDate, @EndDate) - DATEDIFF(week, DATEADD(day, 1, @StartDate), DATEADD(day, 1, @EndDate)) - (CASE WHEN DATEPART(WEEKDAY, @StartDate) IN (1, 7) THEN 1 ELSE 0 END) + 1
@caiosm1005，周六到周日返回0，周六到周一返回1，周五到周六返回0。都和我的定义一致。您的代码不会正确累积（例如，周五到周五返回 6，但周一到周一返回 5）

【解决方案5】：

与 DATEDIFF 一样，我不认为结束日期是间隔的一部分。 @StartDate 和@EndDate 之间的（例如）星期日数是“初始”星期一和@EndDate 之间的星期日数减去这个“初始”星期一和@StartDate 之间的星期日数。知道了这一点，我们可以如下计算工作日数：

DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = '2018/01/01'
SET @EndDate = '2019/01/01'

SELECT DATEDIFF(Day, @StartDate, @EndDate) -- Total Days
  - (DATEDIFF(Day, 0, @EndDate)/7 - DATEDIFF(Day, 0, @StartDate)/7) -- Sundays
  - (DATEDIFF(Day, -1, @EndDate)/7 - DATEDIFF(Day, -1, @StartDate)/7) -- Saturdays

最好的问候！

【讨论】：

完美！这就是我一直在寻找的。特别感谢！

【解决方案6】：

一种方法是结合 case 表达式从头到尾“遍历日期”，该表达式检查当天是不是星期六或星期日并标记它（1 表示工作日，0 表示周末）。最后，只需对标志求和（它将等于 1 标志的计数，因为另一个标志为 0）为您提供工作日数。

您可以使用 GetNums(startNumber,endNumber) 类型的实用程序函数，该函数生成一系列数字，用于从开始日期到结束日期“循环”。参考http://tsql.solidq.com/SourceCodes/GetNums.txt 的实现。逻辑也可以扩展以迎合假期（例如，如果您有假期表）

declare @date1 as datetime = '19900101'
declare @date2 as datetime = '19900120'

select  sum(case when DATENAME(DW,currentDate) not in ('Saturday', 'Sunday') then 1 else 0 end) as noOfWorkDays
from dbo.GetNums(0,DATEDIFF(day,@date1, @date2)-1) as Num
cross apply (select DATEADD(day,n,@date1)) as Dates(currentDate)

【讨论】：

【解决方案7】：

我发现下面的 TSQL 是一个相当优雅的解决方案（我没有运行函数的权限）。我发现DATEDIFF 忽略了DATEFIRST，我希望我一周的第一天是星期一。我还希望将第一个工作日设置为零，如果星期一是周末，则设置为零。这可能会对要求略有不同的人有所帮助:)

它不处理银行假期

SET DATEFIRST 1
SELECT
,(DATEDIFF(DD,  [StartDate], [EndDate]))        
-(DATEDIFF(wk,  [StartDate], [EndDate]))        
-(DATEDIFF(wk, DATEADD(dd,-@@DATEFIRST,[StartDate]), DATEADD(dd,-@@DATEFIRST,[EndDate]))) AS [WorkingDays] 
FROM /*Your Table*/

【讨论】：

【解决方案8】：

另一种计算工作日的方法是使用 WHILE 循环，该循环基本上会遍历一个日期范围，并在发现日期在周一至周五之间时将其加 1。使用 WHILE 循环计算工作日的完整脚本如下所示：

CREATE FUNCTION [dbo].[fn_GetTotalWorkingDaysUsingLoop]
(@DateFrom DATE,
@DateTo   DATE
)
RETURNS INT
AS
     BEGIN
         DECLARE @TotWorkingDays INT= 0;
         WHILE @DateFrom <= @DateTo
             BEGIN
                 IF DATENAME(WEEKDAY, @DateFrom) IN('Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday')
                     BEGIN
                         SET @TotWorkingDays = @TotWorkingDays + 1;
                 END;
                 SET @DateFrom = DATEADD(DAY, 1, @DateFrom);
             END;
         RETURN @TotWorkingDays;
     END;
GO

虽然 WHILE 循环选项更简洁且使用的代码行数更少，但它有可能成为您环境中的性能瓶颈，尤其是当您的日期范围跨越数年时。

您可以在本文中看到更多关于如何计算工作日和小时数的方法： https://www.sqlshack.com/how-to-calculate-work-days-and-hours-in-sql-server/

【讨论】：

【解决方案9】：

Create Function dbo.DateDiff_WeekDays 
(
@StartDate  DateTime,
@EndDate    DateTime
)
Returns Int
As

Begin   

Declare @Result Int = 0

While   @StartDate <= @EndDate
Begin 
    If DateName(DW, @StartDate) not in ('Saturday','Sunday')
        Begin
            Set @Result = @Result +1
        End
        Set @StartDate = DateAdd(Day, +1, @StartDate)
End

Return @Result

结束

【讨论】：

【解决方案10】：

创建函数如：

CREATE FUNCTION dbo.fn_WorkDays(@StartDate DATETIME, @EndDate DATETIME= NULL )
RETURNS INT 
AS
BEGIN
       DECLARE @Days int
       SET @Days = 0

       IF @EndDate = NULL
              SET @EndDate = EOMONTH(@StartDate) --last date of the month

       WHILE DATEDIFF(dd,@StartDate,@EndDate) >= 0
       BEGIN
              IF DATENAME(dw, @StartDate) <> 'Saturday' 
                     and DATENAME(dw, @StartDate) <> 'Sunday' 
                     and Not ((Day(@StartDate) = 1 And Month(@StartDate) = 1)) --New Year's Day.
                     and Not ((Day(@StartDate) = 4 And Month(@StartDate) = 7)) --Independence Day.
              BEGIN
                     SET @Days = @Days + 1
              END

              SET @StartDate = DATEADD(dd,1,@StartDate)
       END

       RETURN  @Days
END

你可以像这样调用函数：

select dbo.fn_WorkDays('1/1/2016', '9/25/2016')

或者喜欢：

select dbo.fn_WorkDays(StartDate, EndDate) 
from table1

【讨论】：

【解决方案11】：

我接受的答案版本是使用DATEPART 的函数，所以我不必在行上进行字符串比较

DATENAME(dw, @StartDate) = 'Sunday'

无论如何，这是我的业务 datediff 函数

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO

CREATE FUNCTION BDATEDIFF
(
    @startdate as DATETIME,
    @enddate as DATETIME
)
RETURNS INT
AS
BEGIN
    DECLARE @res int

SET @res = (DATEDIFF(dd, @startdate, @enddate) + 1)
    -(DATEDIFF(wk, @startdate, @enddate) * 2)
    -(CASE WHEN DATEPART(dw, @startdate) = 1 THEN 1 ELSE 0 END)
    -(CASE WHEN DATEPART(dw, @enddate) = 7 THEN 1 ELSE 0 END)

    RETURN @res
END
GO

【讨论】：

【解决方案12】：

这对我有用，在我的国家，周六和周日是非工作日。

@StartDate 和@EndDate 的时间对我来说很重要。

CREATE FUNCTION [dbo].[fnGetCountWorkingBusinessDays]
(
    @StartDate as DATETIME,
    @EndDate as DATETIME
)
RETURNS INT
AS
BEGIN
    DECLARE @res int

SET @StartDate = CASE 
    WHEN DATENAME(dw, @StartDate) = 'Saturday' THEN DATEADD(dd, 2, DATEDIFF(dd, 0, @StartDate))
    WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN DATEADD(dd, 1, DATEDIFF(dd, 0, @StartDate))
    ELSE @StartDate END

SET @EndDate = CASE 
    WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN DATEADD(dd, 0, DATEDIFF(dd, 0, @EndDate))
    WHEN DATENAME(dw, @EndDate) = 'Sunday' THEN DATEADD(dd, -1, DATEDIFF(dd, 0, @EndDate))
    ELSE @EndDate END


SET @res =
    (DATEDIFF(hour, @StartDate, @EndDate) / 24)
  - (DATEDIFF(wk, @StartDate, @EndDate) * 2)

SET @res = CASE WHEN @res < 0 THEN 0 ELSE @res END

    RETURN @res
END

GO

【讨论】：

【解决方案13】：

使用日期表：

    DECLARE 
        @StartDate date = '2014-01-01',
        @EndDate date = '2014-01-31'; 
    SELECT 
        COUNT(*) As NumberOfWeekDays
    FROM dbo.Calendar
    WHERE CalendarDate BETWEEN @StartDate AND @EndDate
      AND IsWorkDay = 1;

如果没有，可以使用数字表：

    DECLARE 
    @StartDate datetime = '2014-01-01',
    @EndDate datetime = '2014-01-31'; 
    SELECT 
    SUM(CASE WHEN DATEPART(dw, DATEADD(dd, Number-1, @StartDate)) BETWEEN 2 AND 6 THEN 1 ELSE 0 END) As NumberOfWeekDays
    FROM dbo.Numbers
    WHERE Number <= DATEDIFF(dd, @StartDate, @EndDate) + 1 -- Number table starts at 1, we want a 0 base

它们都应该很快，并且消除了歧义/复杂性。第一个选项是最好的，但如果您没有日历表，您总是可以创建一个带有 CTE 的数字表。

【讨论】：

【解决方案14】：

如果您需要将工作日添加到给定日期，您可以创建一个依赖于日历表的函数，如下所述：

CREATE TABLE Calendar
(
  dt SMALLDATETIME PRIMARY KEY, 
  IsWorkDay BIT
);

--fill the rows with normal days, weekends and holidays.


create function AddWorkingDays (@initialDate smalldatetime, @numberOfDays int)
    returns smalldatetime as 

    begin
        declare @result smalldatetime
        set @result = 
        (
            select t.dt from
            (
                select dt, ROW_NUMBER() over (order by dt) as daysAhead from calendar 
                where dt > @initialDate
                and IsWorkDay = 1
                ) t
            where t.daysAhead = @numberOfDays
        )

        return @result
    end

【讨论】：

+1 我最终使用了a similar solution here

【解决方案15】：

感谢 Bogdan Maxim 和 Peter Mortensen。这是他们的帖子，我只是在函数中添加了假期（假设您有一个带有日期时间字段“HolDate”的表“tblHolidays”。

--Changing current database to the Master database allows function to be shared by everyone.
USE MASTER
GO
--If the function already exists, drop it.
IF EXISTS
(
    SELECT *
    FROM dbo.SYSOBJECTS
    WHERE ID = OBJECT_ID(N'[dbo].[fn_WorkDays]')
    AND XType IN (N'FN', N'IF', N'TF')
)

DROP FUNCTION [dbo].[fn_WorkDays]
GO
 CREATE FUNCTION dbo.fn_WorkDays
--Presets
--Define the input parameters (OK if reversed by mistake).
(
    @StartDate DATETIME,
    @EndDate   DATETIME = NULL --@EndDate replaced by @StartDate when DEFAULTed
)

--Define the output data type.
RETURNS INT

AS
--Calculate the RETURN of the function.
BEGIN
    --Declare local variables
    --Temporarily holds @EndDate during date reversal.
    DECLARE @Swap DATETIME

    --If the Start Date is null, return a NULL and exit.
    IF @StartDate IS NULL
        RETURN NULL

    --If the End Date is null, populate with Start Date value so will have two dates (required by DATEDIFF below).
    IF @EndDate IS NULL
        SELECT @EndDate = @StartDate

    --Strip the time element from both dates (just to be safe) by converting to whole days and back to a date.
    --Usually faster than CONVERT.
    --0 is a date (01/01/1900 00:00:00.000)
    SELECT @StartDate = DATEADD(dd,DATEDIFF(dd,0,@StartDate), 0),
            @EndDate   = DATEADD(dd,DATEDIFF(dd,0,@EndDate)  , 0)

    --If the inputs are in the wrong order, reverse them.
    IF @StartDate > @EndDate
        SELECT @Swap      = @EndDate,
               @EndDate   = @StartDate,
               @StartDate = @Swap

    --Calculate and return the number of workdays using the input parameters.
    --This is the meat of the function.
    --This is really just one formula with a couple of parts that are listed on separate lines for documentation purposes.
    RETURN (
        SELECT
        --Start with total number of days including weekends
        (DATEDIFF(dd,@StartDate, @EndDate)+1)
        --Subtact 2 days for each full weekend
        -(DATEDIFF(wk,@StartDate, @EndDate)*2)
        --If StartDate is a Sunday, Subtract 1
        -(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday'
            THEN 1
            ELSE 0
        END)
        --If EndDate is a Saturday, Subtract 1
        -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday'
            THEN 1
            ELSE 0
        END)
        --Subtract all holidays
        -(Select Count(*) from [DB04\DB04].[Gateway].[dbo].[tblHolidays]
          where  [HolDate] between @StartDate and @EndDate )
        )
    END  
GO
-- Test Script
/*
declare @EndDate datetime= dateadd(m,2,getdate())
print @EndDate
select  [Master].[dbo].[fn_WorkDays] (getdate(), @EndDate)
*/

【讨论】：

嗨 Dan B。只是为了让您知道您的版本假定表 tblHolidays 不包含星期六和星期一，这有时会发生。无论如何，感谢您分享您的版本。干杯
Julio - 是的 - 我的版本确实假设周六和周日（不是周一）是周末，因此不是“非工作日”。但是，如果您在周末工作，那么我猜每天都是“工作日”，您可以注释掉该子句的周六和周日部分，然后将所有假期添加到 tblHolidays 表中。
谢谢丹。我将它合并到我的函数中，添加一个周末检查，因为我的 DateDimensions 表包括所有日期、假期等。使用你的函数，我刚刚添加：并且 IsWeekend = 0 在 [HolDate] 之间的 StartDate 和 EndDate 之后）
如果Holiday 表包含周末的假期，您可以像这样修改条件：WHERE HolDate BETWEEN @StartDate AND @EndDate AND DATEPART(dw, HolDate) BETWEEN 2 AND 6 以仅计算周一至周五的假期。

【解决方案16】：

这是一个运行良好的版本（我认为）。假期表包含 Holiday_date 列，其中包含贵公司遵守的假期。

DECLARE @RAWDAYS INT

   SELECT @RAWDAYS =  DATEDIFF(day, @StartDate, @EndDate )--+1
                    -( 2 * DATEDIFF( week, @StartDate, @EndDate ) )
                    + CASE WHEN DATENAME(dw, @StartDate) = 'Saturday' THEN 1 ELSE 0 END
                    - CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END 

   SELECT  @RAWDAYS - COUNT(*) 
     FROM HOLIDAY NumberOfBusinessDays
    WHERE [Holiday_Date] BETWEEN @StartDate+1 AND @EndDate

【讨论】：

那些假期日期也可能落在周末。对于一些人来说，周日的假期将被下周一取代。

【解决方案17】：

我在这里举了各种例子，但在我的特殊情况下，我们有一个用于交付的@PromisedDate 和一个用于实际接收物品的@ReceivedDate。当在“PromisedDate”之前收到一个项目时，除非我按日历顺序订购传递给函数的日期，否则计算的总计不正确。不想每次都检查日期，我更改了函数来为我处理这个问题。

Create FUNCTION [dbo].[fnGetBusinessDays]
(
 @PromiseDate date,
 @ReceivedDate date
)
RETURNS integer
AS
BEGIN
 DECLARE @days integer

 SELECT @days = 
    Case when @PromiseDate > @ReceivedDate Then
        DATEDIFF(d,@PromiseDate,@ReceivedDate) + 
        ABS(DATEDIFF(wk,@PromiseDate,@ReceivedDate)) * 2 +
        CASE 
            WHEN DATENAME(dw, @PromiseDate) <> 'Saturday' AND DATENAME(dw, @ReceivedDate) = 'Saturday' THEN 1 
            WHEN DATENAME(dw, @PromiseDate) = 'Saturday' AND DATENAME(dw, @ReceivedDate) <> 'Saturday' THEN -1 
            ELSE 0
        END +
        (Select COUNT(*) FROM CompanyHolidays 
            WHERE HolidayDate BETWEEN @ReceivedDate AND @PromiseDate 
            AND DATENAME(dw, HolidayDate) <> 'Saturday' AND DATENAME(dw, HolidayDate) <> 'Sunday')
    Else
        DATEDIFF(d,@PromiseDate,@ReceivedDate)  -
        ABS(DATEDIFF(wk,@PromiseDate,@ReceivedDate)) * 2  -
            CASE 
                WHEN DATENAME(dw, @PromiseDate) <> 'Saturday' AND DATENAME(dw, @ReceivedDate) = 'Saturday' THEN 1 
                WHEN DATENAME(dw, @PromiseDate) = 'Saturday' AND DATENAME(dw, @ReceivedDate) <> 'Saturday' THEN -1 
                ELSE 0
            END -
        (Select COUNT(*) FROM CompanyHolidays 
            WHERE HolidayDate BETWEEN @PromiseDate and @ReceivedDate 
            AND DATENAME(dw, HolidayDate) <> 'Saturday' AND DATENAME(dw, HolidayDate) <> 'Sunday')
    End


 RETURN (@days)

END

【讨论】：

【解决方案18】：

对于包括假期在内的日期之间的差异，我这样做了：

1) 假期表：

    CREATE TABLE [dbo].[Holiday](
[Id] [int] IDENTITY(1,1) NOT NULL,
[Name] [nvarchar](50) NULL,
[Date] [datetime] NOT NULL)

2) 我有这样的计划表，想填充空的 Work_Days 列：

    CREATE TABLE [dbo].[Plan_Phase](
[Id] [int] IDENTITY(1,1) NOT NULL,
[Id_Plan] [int] NOT NULL,
[Id_Phase] [int] NOT NULL,
[Start_Date] [datetime] NULL,
[End_Date] [datetime] NULL,
[Work_Days] [int] NULL)

3) 因此，为了让“Work_Days”稍后填写我的专栏，只需：

SELECT Start_Date, End_Date,
 (DATEDIFF(dd, Start_Date, End_Date) + 1)
-(DATEDIFF(wk, Start_Date, End_Date) * 2)
-(SELECT COUNT(*) From Holiday Where Date  >= Start_Date AND Date <= End_Date)
-(CASE WHEN DATENAME(dw, Start_Date) = 'Sunday' THEN 1 ELSE 0 END)
-(CASE WHEN DATENAME(dw, End_Date) = 'Saturday' THEN 1 ELSE 0 END)
-(CASE WHEN (SELECT COUNT(*) From Holiday Where Start_Date  = Date) > 0 THEN 1 ELSE 0 END)
-(CASE WHEN (SELECT COUNT(*) From Holiday Where End_Date  = Date) > 0 THEN 1 ELSE 0 END) AS Work_Days
from Plan_Phase

希望我能帮上忙。

干杯

【讨论】：

关于你的假期减法。如果开始日期是 1 月 1 日，结束日期是 12 月 31 日怎么办？你只会减去 2 - 这是错误的。我建议对 End_Date 使用 DATEDIFF(day, Start_Date, Date) 而不是整个 'SELECT COUNT(*) FROM Holiday ...'。

【解决方案19】：

（我比评论权限差了几分）

如果您决定放弃 CMS's elegant solution 中的 +1 天，请注意，如果您的开始日期和结束日期在同一个周末，您会得到否定的答案。即，2008/10/26 到 2008/10/26 返回 -1。

我比较简单的解决方案：

select @Result = (..CMS's answer..)
if  (@Result < 0)
        select @Result = 0
    RETURN @Result

.. 它还将 start date 在 end date 之后的所有错误帖子设置为零。您可能会或可能不会寻找的东西。

【讨论】：

【解决方案20】：

CREATE FUNCTION x
(
    @StartDate DATETIME,
    @EndDate DATETIME
)
RETURNS INT
AS
BEGIN
    DECLARE @Teller INT

    SET @StartDate = DATEADD(dd,1,@StartDate)

    SET @Teller = 0
    IF DATEDIFF(dd,@StartDate,@EndDate) <= 0
    BEGIN
        SET @Teller = 0 
    END
    ELSE
    BEGIN
        WHILE
            DATEDIFF(dd,@StartDate,@EndDate) >= 0
        BEGIN
            IF DATEPART(dw,@StartDate) < 6
            BEGIN
                SET @Teller = @Teller + 1
            END
            SET @StartDate = DATEADD(dd,1,@StartDate)
        END
    END
    RETURN @Teller
END

【讨论】：

【解决方案21】：

在 Calculating Work Days 你可以找到一篇关于这个主题的好文章，但是你可以看到它不是那么先进。

--Changing current database to the Master database allows function to be shared by everyone.
USE MASTER
GO
--If the function already exists, drop it.
IF EXISTS
(
    SELECT *
    FROM dbo.SYSOBJECTS
    WHERE ID = OBJECT_ID(N'[dbo].[fn_WorkDays]')
    AND XType IN (N'FN', N'IF', N'TF')
)
DROP FUNCTION [dbo].[fn_WorkDays]
GO
 CREATE FUNCTION dbo.fn_WorkDays
--Presets
--Define the input parameters (OK if reversed by mistake).
(
    @StartDate DATETIME,
    @EndDate   DATETIME = NULL --@EndDate replaced by @StartDate when DEFAULTed
)

--Define the output data type.
RETURNS INT

AS
--Calculate the RETURN of the function.
BEGIN
    --Declare local variables
    --Temporarily holds @EndDate during date reversal.
    DECLARE @Swap DATETIME

    --If the Start Date is null, return a NULL and exit.
    IF @StartDate IS NULL
        RETURN NULL

    --If the End Date is null, populate with Start Date value so will have two dates (required by DATEDIFF below).
     IF @EndDate IS NULL
        SELECT @EndDate = @StartDate

    --Strip the time element from both dates (just to be safe) by converting to whole days and back to a date.
    --Usually faster than CONVERT.
    --0 is a date (01/01/1900 00:00:00.000)
     SELECT @StartDate = DATEADD(dd,DATEDIFF(dd,0,@StartDate), 0),
            @EndDate   = DATEADD(dd,DATEDIFF(dd,0,@EndDate)  , 0)

    --If the inputs are in the wrong order, reverse them.
     IF @StartDate > @EndDate
        SELECT @Swap      = @EndDate,
               @EndDate   = @StartDate,
               @StartDate = @Swap

    --Calculate and return the number of workdays using the input parameters.
    --This is the meat of the function.
    --This is really just one formula with a couple of parts that are listed on separate lines for documentation purposes.
     RETURN (
        SELECT
        --Start with total number of days including weekends
        (DATEDIFF(dd,@StartDate, @EndDate)+1)
        --Subtact 2 days for each full weekend
        -(DATEDIFF(wk,@StartDate, @EndDate)*2)
        --If StartDate is a Sunday, Subtract 1
        -(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday'
            THEN 1
            ELSE 0
        END)
        --If EndDate is a Saturday, Subtract 1
        -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday'
            THEN 1
            ELSE 0
        END)
        )
    END
GO

如果您需要使用自定义日历，您可能需要添加一些检查和一些参数。希望它能提供一个好的起点。

【讨论】：

感谢您提供链接以了解其工作原理。 sqlservercentral 写的很棒！

【解决方案22】：

DECLARE @StartDate datetime,@EndDate datetime

select @StartDate='3/2/2010', @EndDate='3/7/2010'

DECLARE @TotalDays INT,@WorkDays INT

DECLARE @ReducedDayswithEndDate INT

DECLARE @WeekPart INT

DECLARE @DatePart INT

SET @TotalDays= DATEDIFF(day, @StartDate, @EndDate) +1

SELECT @ReducedDayswithEndDate = CASE DATENAME(weekday, @EndDate)
    WHEN 'Saturday' THEN 1
    WHEN 'Sunday' THEN 2
    ELSE 0 END

SET @TotalDays=@TotalDays-@ReducedDayswithEndDate

SET @WeekPart=@TotalDays/7;

SET @DatePart=@TotalDays%7;

SET @WorkDays=(@WeekPart*5)+@DatePart

SELECT @WorkDays

【讨论】：

如果您要使用函数，最好使用基于表的函数，如the answer by Mário Meyrelles

【解决方案23】：

 DECLARE @TotalDays INT,@WorkDays INT
 DECLARE @ReducedDayswithEndDate INT
 DECLARE @WeekPart INT
 DECLARE @DatePart INT

 SET @TotalDays= DATEDIFF(day, @StartDate, @EndDate) +1
 SELECT @ReducedDayswithEndDate = CASE DATENAME(weekday, @EndDate)
  WHEN 'Saturday' THEN 1
  WHEN 'Sunday' THEN 2
  ELSE 0 END 
 SET @TotalDays=@TotalDays-@ReducedDayswithEndDate
 SET @WeekPart=@TotalDays/7;
 SET @DatePart=@TotalDays%7;
 SET @WorkDays=(@WeekPart*5)+@DatePart

 RETURN @WorkDays

【讨论】：

如果您发布代码、XML 或数据示例，请在文本编辑器中突出显示这些行，然后单击编辑器工具栏上的“代码示例”按钮 ({})很好地格式化和语法高亮它！
太好了，不需要外围功能或使用它来更新数据库。谢谢。顺便说一句，喜欢saltire :-)
超级解决方案。我在 webi Universe 中使用变量的公式来计算 2 个表列中的日期之间的工作日 (MF)，如下所示 ...((((DATEDIFF(day, table.col1, table.col2) +1)- ((CASE DATENAME(weekday, table.col2) WHEN 'Saturday' THEN 1 WHEN 'Sunday' THEN 2 ELSE 0 END )))/7)*5)+(((DATEDIFF(day, table.col1, table.col2) ) +1)-((CASE DATENAME(weekday, table.col2) WHEN 'Saturday' THEN 1 WHEN 'Sunday' THEN 2 ELSE 0 END )))%7)

【解决方案24】：

对于周一至周五的工作日，您可以使用单个 SELECT 来完成，如下所示：

DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = '2008/10/01'
SET @EndDate = '2008/10/31'


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)

如果你想包括假期，你必须稍微考虑一下......

【讨论】：

我刚刚意识到这段代码并不总是有效！我试过这个： SET @StartDate = '28-mar-2011' SET @EndDate = '29-mar-2011' 答案它算作 2 天
@greektreat 它工作正常。只是@StartDate 和@EndDate 都包含在计数中。如果您希望星期一到星期二算作 1 天，只需删除第一个 DATEDIFF 后的“+1”即可。然后你也会得到 Fri->Sat=0, Fri->Sun=0, Fri->Mon=1。
作为@JoeDaley 的后续行动。当您删除 DATEDIFF 后的 + 1 以从计数中排除 startdate 时，您还需要调整其中的 CASE 部分。我最终使用了这个： +(CASE WHEN DATENAME(dw, @StartDate) = 'Saturday' THEN 1 ELSE 0 END) - (CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END)跨度>
datename 函数取决于区域设置。一个更强大但也更模糊的解决方案是将最后两行替换为：-(case datepart(dw, @StartDate)+@@datefirst when 8 then 1 else 0 end) -(case datepart(dw, @EndDate)+@@datefirst when 7 then 1 when 14 then 1 else 0 end)
为了澄清@Sequenzia 的评论，您将完全删除有关星期日的案例陈述，只留下+(CASE WHEN DATENAME(dw, @StartDate) = 'Saturday' THEN 1 ELSE 0 END) - (CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END)