如何计算两个日期之间的营业时间？ 例如，我们有两个日期; 01/01/2010 15:00和04/01/2010 12:00 我们工作时间的工作时间为09:00至17:00 如何用sql计算工作时间？
答案 0 :(得分：26)
Baran的答案针对SQL 2005进行了修复和修改
SQL 2008及更高版本：
-- =============================================
-- Author: Baran Kaynak (modified by Kodak 2012-04-18)
-- Create date: 14.03.2011
-- Description: 09:30 ile 17:30 arasındaki iş saatlerini hafta sonlarını almayarak toplar.
-- =============================================
CREATE FUNCTION [dbo].[WorkTime]
(
@StartDate DATETIME,
@FinishDate DATETIME
)
RETURNS BIGINT
AS
BEGIN
DECLARE @Temp BIGINT
SET @Temp=0
DECLARE @FirstDay DATE
SET @FirstDay = CONVERT(DATE, @StartDate, 112)
DECLARE @LastDay DATE
SET @LastDay = CONVERT(DATE, @FinishDate, 112)
DECLARE @StartTime TIME
SET @StartTime = CONVERT(TIME, @StartDate)
DECLARE @FinishTime TIME
SET @FinishTime = CONVERT(TIME, @FinishDate)
DECLARE @WorkStart TIME
SET @WorkStart = '09:00'
DECLARE @WorkFinish TIME
SET @WorkFinish = '17:00'
DECLARE @DailyWorkTime BIGINT
SET @DailyWorkTime = DATEDIFF(MINUTE, @WorkStart, @WorkFinish)
IF (@StartTime<@WorkStart)
BEGIN
SET @StartTime = @WorkStart
END
IF (@FinishTime>@WorkFinish)
BEGIN
SET @FinishTime=@WorkFinish
END
IF (@FinishTime<@WorkStart)
BEGIN
SET @FinishTime=@WorkStart
END
IF (@StartTime>@WorkFinish)
BEGIN
SET @StartTime = @WorkFinish
END
DECLARE @CurrentDate DATE
SET @CurrentDate = @FirstDay
DECLARE @LastDate DATE
SET @LastDate = @LastDay
WHILE(@CurrentDate<=@LastDate)
BEGIN
IF (DATEPART(dw, @CurrentDate)!=1 AND DATEPART(dw, @CurrentDate)!=7)
BEGIN
IF (@CurrentDate!=@FirstDay) AND (@CurrentDate!=@LastDay)
BEGIN
SET @Temp = @Temp + @DailyWorkTime
END
--IF it starts at startdate and it finishes not this date find diff between work finish and start as minutes
ELSE IF (@CurrentDate=@FirstDay) AND (@CurrentDate!=@LastDay)
BEGIN
SET @Temp = @Temp + DATEDIFF(MINUTE, @StartTime, @WorkFinish)
END
ELSE IF (@CurrentDate!=@FirstDay) AND (@CurrentDate=@LastDay)
BEGIN
SET @Temp = @Temp + DATEDIFF(MINUTE, @WorkStart, @FinishTime)
END
--IF it starts and finishes in the same date
ELSE IF (@CurrentDate=@FirstDay) AND (@CurrentDate=@LastDay)
BEGIN
SET @Temp = DATEDIFF(MINUTE, @StartTime, @FinishTime)
END
END
SET @CurrentDate = DATEADD(day, 1, @CurrentDate)
END
-- Return the result of the function
IF @Temp<0
BEGIN
SET @Temp=0
END
RETURN @Temp
END
SQL 2005及以下版本：
-- =============================================
-- Author: Baran Kaynak (modified by Kodak 2012-04-18)
-- Create date: 14.03.2011
-- Description: 09:30 ile 17:30 arasındaki iş saatlerini hafta sonlarını almayarak toplar.
-- =============================================
CREATE FUNCTION [dbo].[WorkTime]
(
@StartDate DATETIME,
@FinishDate DATETIME
)
RETURNS BIGINT
AS
BEGIN
DECLARE @Temp BIGINT
SET @Temp=0
DECLARE @FirstDay DATETIME
SET @FirstDay = DATEADD(dd, 0, DATEDIFF(dd, 0, @StartDate))
DECLARE @LastDay DATETIME
SET @LastDay = DATEADD(dd, 0, DATEDIFF(dd, 0, @FinishDate))
DECLARE @StartTime DATETIME
SET @StartTime = @StartDate - DATEADD(dd, DATEDIFF(dd, 0, @StartDate), 0)
DECLARE @FinishTime DATETIME
SET @FinishTime = @FinishDate - DATEADD(dd, DATEDIFF(dd, 0, @FinishDate), 0)
DECLARE @WorkStart DATETIME
SET @WorkStart = CONVERT(DATETIME, '09:00', 8)
DECLARE @WorkFinish DATETIME
SET @WorkFinish = CONVERT(DATETIME, '17:00', 8)
DECLARE @DailyWorkTime BIGINT
SET @DailyWorkTime = DATEDIFF(MINUTE, @WorkStart, @WorkFinish)
IF (@StartTime<@WorkStart)
BEGIN
SET @StartTime = @WorkStart
END
IF (@FinishTime>@WorkFinish)
BEGIN
SET @FinishTime=@WorkFinish
END
IF (@FinishTime<@WorkStart)
BEGIN
SET @FinishTime=@WorkStart
END
IF (@StartTime>@WorkFinish)
BEGIN
SET @StartTime = @WorkFinish
END
DECLARE @CurrentDate DATETIME
SET @CurrentDate = @FirstDay
DECLARE @LastDate DATETIME
SET @LastDate = @LastDay
WHILE(@CurrentDate<=@LastDate)
BEGIN
IF (DATEPART(dw, @CurrentDate)!=1 AND DATEPART(dw, @CurrentDate)!=7)
BEGIN
IF (@CurrentDate!=@FirstDay) AND (@CurrentDate!=@LastDay)
BEGIN
SET @Temp = @Temp + @DailyWorkTime
END
--IF it starts at startdate and it finishes not this date find diff between work finish and start as minutes
ELSE IF (@CurrentDate=@FirstDay) AND (@CurrentDate!=@LastDay)
BEGIN
SET @Temp = @Temp + DATEDIFF(MINUTE, @StartTime, @WorkFinish)
END
ELSE IF (@CurrentDate!=@FirstDay) AND (@CurrentDate=@LastDay)
BEGIN
SET @Temp = @Temp + DATEDIFF(MINUTE, @WorkStart, @FinishTime)
END
--IF it starts and finishes in the same date
ELSE IF (@CurrentDate=@FirstDay) AND (@CurrentDate=@LastDay)
BEGIN
SET @Temp = DATEDIFF(MINUTE, @StartTime, @FinishTime)
END
END
SET @CurrentDate = DATEADD(day, 1, @CurrentDate)
END
-- Return the result of the function
IF @Temp<0
BEGIN
SET @Temp=0
END
RETURN @Temp
END
答案 1 :(得分：19)
我知道这篇帖子已经很老了，但这是我最近编写的一个函数来计算任意两个事件之间的营业时间/分钟。它还考虑了必须在表格中定义的任何假期。
该函数以分钟为单位返回间隔 - 您可以除以60以获得所需的小时数。
这已经在SQL Server 2008上进行了测试。希望它对某人有帮助。
Create Function GetWorkingMin(@StartDate DateTime, @EndDate DateTime, @Country Varchar(2)) Returns Int
AS
Begin
Declare @WorkMin int = 0 -- Initialize counter
Declare @Reverse bit -- Flag to hold if direction is reverse
Declare @StartHour int = 9 -- Start of business hours (can be supplied as an argument if needed)
Declare @EndHour int = 17 -- End of business hours (can be supplied as an argument if needed)
Declare @Holidays Table (HDate DateTime) -- Table variable to hold holidayes
-- If dates are in reverse order, switch them and set flag
If @StartDate>@EndDate
Begin
Declare @TempDate DateTime=@StartDate
Set @StartDate=@EndDate
Set @EndDate=@TempDate
Set @Reverse=1
End
Else Set @Reverse = 0
-- Get country holidays from table based on the country code (Feel free to remove this or modify as per your DB schema)
Insert Into @Holidays (HDate) Select HDate from HOLIDAY Where COUNTRYCODE=@Country and HDATE>=DateAdd(dd, DateDiff(dd,0,@StartDate), 0)
If DatePart(HH, @StartDate)<@StartHour Set @StartDate = DateAdd(hour, @StartHour, DateDiff(DAY, 0, @StartDate)) -- If Start time is less than start hour, set it to start hour
If DatePart(HH, @StartDate)>=@EndHour+1 Set @StartDate = DateAdd(hour, @StartHour+24, DateDiff(DAY, 0, @StartDate)) -- If Start time is after end hour, set it to start hour of next day
If DatePart(HH, @EndDate)>=@EndHour+1 Set @EndDate = DateAdd(hour, @EndHour, DateDiff(DAY, 0, @EndDate)) -- If End time is after end hour, set it to end hour
If DatePart(HH, @EndDate)<@StartHour Set @EndDate = DateAdd(hour, @EndHour-24, DateDiff(DAY, 0, @EndDate)) -- If End time is before start hour, set it to end hour of previous day
If @StartDate>@EndDate Return 0
-- If Start and End is on same day
If DateDiff(Day,@StartDate,@EndDate) <= 0
Begin
If Datepart(dw,@StartDate)>1 And DATEPART(dw,@StartDate)<7 -- If day is between sunday and saturday
If (Select Count(*) From @Holidays Where HDATE=DateAdd(dd, DateDiff(dd,0,@StartDate), 0)) = 0 -- If day is not a holiday
If @EndDate<@StartDate Return 0 Else Set @WorkMin=DATEDIFF(MI, @StartDate, @EndDate) -- Calculate difference
Else Return 0
Else Return 0
End
Else Begin
Declare @Partial int=1 -- Set partial day flag
While DateDiff(Day,@StartDate,@EndDate) > 0 -- While start and end days are different
Begin
If Datepart(dw,@StartDate)>1 And DATEPART(dw,@StartDate)<7 -- If this is a weekday
Begin
If (Select Count(*) From @Holidays Where HDATE=DateAdd(dd, DateDiff(dd,0,@StartDate), 0)) = 0 -- If this is not a holiday
Begin
If @Partial=1 -- If this is the first iteration, calculate partial time
Begin
Set @WorkMin=@WorkMin + DATEDIFF(MI, @StartDate, DateAdd(hour, @EndHour, DateDiff(DAY, 0, @StartDate)))
Set @StartDate=DateAdd(hour, @StartHour+24, DateDiff(DAY, 0, @StartDate))
Set @Partial=0
End
Else Begin -- If this is a full day, add full minutes
Set @WorkMin=@WorkMin + (@EndHour-@StartHour)*60
Set @StartDate = DATEADD(DD,1,@StartDate)
End
End
Else Set @StartDate = DATEADD(DD,1,@StartDate)
End
Else Set @StartDate = DATEADD(DD,1,@StartDate)
End
If Datepart(dw,@StartDate)>1 And DATEPART(dw,@StartDate)<7 -- If last day is a weekday
If (Select Count(*) From @Holidays Where HDATE=DateAdd(dd, DateDiff(dd,0,@StartDate), 0)) = 0 -- And it is not a holiday
If @Partial=0 Set @WorkMin=@WorkMin + DATEDIFF(MI, @StartDate, @EndDate) Else Set @WorkMin=@WorkMin + DATEDIFF(MI, DateAdd(hour, @StartHour, DateDiff(DAY, 0, @StartDate)), @EndDate)
End
If @Reverse=1 Set @WorkMin=-@WorkMin
Return @WorkMin
End
答案 2 :(得分：8)
第一步是计算工作日，如下面的脚本所示：
DECLARE @TotalWorkDays INT, @TotalTimeDiff DECIMAL(18, 2), @DateFrom DATETIME, @DateTo DATETIME;
SET @DateFrom = '2017-06-05 11:19:11.287';
SET @DateTo = '2017-06-07 09:53:14.750';
SET @TotalWorkDays = DATEDIFF(DAY, @DateFrom, @DateTo)
-(DATEDIFF(WEEK, @DateFrom, @DateTo) * 2)
-CASE
WHEN DATENAME(WEEKDAY, @DateFrom) = 'Sunday'
THEN 1
ELSE 0
END+CASE
WHEN DATENAME(WEEKDAY, @DateTo) = 'Saturday'
THEN 1
ELSE 0
END;
第二步涉及在两个日期之间获得秒数的差异，并将该差异转换为小时数除以3600.0，如下面的脚本所示：
SET @TotalTimeDiff =
(
SELECT DATEDIFF(SECOND,
(
SELECT CONVERT(TIME, @DateFrom)
),
(
SELECT CONVERT(TIME, @DateTo)
)) / 3600.0
);
最后一部分涉及将上面第一步的输出乘以24（一天中的总小时数），然后将其添加到第二步的输出中：
SELECT(@TotalWorkDays * 24.00) + @TotalTimeDiff;
最后，可用于创建用户定义函数以计算工作时间的完整脚本如下所示：
CREATE FUNCTION [dbo].[fn_GetTotalWorkingHours]
(
@DateFrom Datetime,
@DateTo Datetime
)
RETURNS DECIMAL(18,2)
AS
BEGIN
DECLARE @TotalWorkDays INT, @TotalTimeDiff DECIMAL(18, 2)
SET @TotalWorkDays = DATEDIFF(DAY, @DateFrom, @DateTo)
-(DATEDIFF(WEEK, @DateFrom, @DateTo) * 2)
-CASE
WHEN DATENAME(WEEKDAY, @DateFrom) = 'Sunday'
THEN 1
ELSE 0
END+CASE
WHEN DATENAME(WEEKDAY, @DateTo) = 'Saturday'
THEN 1
ELSE 0
END;
SET @TotalTimeDiff =
(
SELECT DATEDIFF(SECOND,
(
SELECT CONVERT(TIME, @DateFrom)
),
(
SELECT CONVERT(TIME, @DateTo)
)) / 3600.0
);
RETURN (SELECT(@TotalWorkDays * 24.00) + @TotalTimeDiff)
END
GO
本文描述了完整的方法： https://www.sqlshack.com/how-to-calculate-work-days-and-hours-in-sql-server/
答案 3 :(得分：7)
DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
DECLARE @WORKINGHOURS INT
DECLARE @Days INT
SET @StartDate = '2010/01/01'
SET @EndDate = '2010/04/01'
--number of working days
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)
--8 hours a day
SET @WORKINGHOURS = @Days * 8
SELECT @WORKINGHOURS
答案 4 :(得分：7)
来自@ Pavanred的替代解决方案，从更基于数据的角度出发：
创建一个包含您要在其中考虑的所有日期的表。对于每一天，设置一些工作小时，如下：
WorkingDate Hours Comment
=========== ===== ==================
1 Jan 2011 0 Saturday
2 Jan 2011 0 Sunday
3 Jan 2011 0 Public Holiday
4 Jan 2011 8 Normal working day
5 Jan 2011 8 Normal working day
-- and so on, for all the days you want to report on.
这需要进行少量设置 - 您可以自动将其预先填充数周而不是周末，然后根据需要调整公众假期等。
但是，你在设置中失去了什么，你很容易获得查询：
SELECT
SUM(Hours)
FROM
working_days
WHERE
WorkingDate BETWEEN @StartDate AND @EndDate
...如果您需要开始为定义工作日的内容添加更复杂的规则，或者如果您的工作时间因日期而异，等等，这可以作为一种更简单的方法。
它还使规则更容易“可编辑”，因为您无需更改任何实际代码来更改工作日的定义，添加公共假期等。
答案 5 :(得分：5)
-- =============================================
-- Author: Baran Kaynak
-- Create date: 14.03.2011
-- Description: 09:30 ile 17:30 arasındaki iş saatlerini hafta sonlarını almayarak toplar.
-- =============================================
CREATE FUNCTION [dbo].[WorkTime]
(
@StartDate DATETIME,
@FinishDate DATETIME
)
RETURNS BIGINT
AS
BEGIN
DECLARE @Temp BIGINT
SET @Temp=0
DECLARE @FirstDay DATE
SET @FirstDay = CONVERT(DATE, @StartDate, 112)
DECLARE @LastDay DATE
SET @LastDay = CONVERT(DATE, @FinishDate, 112)
DECLARE @StartTime TIME
SET @StartTime = CONVERT(TIME, @StartDate)
DECLARE @FinishTime TIME
SET @FinishTime = CONVERT(TIME, @FinishDate)
DECLARE @WorkStart TIME
SET @WorkStart = '09:30'
DECLARE @WorkFinish TIME
SET @WorkFinish = '17:30'
IF (@StartTime<@WorkStart)
BEGIN
SET @StartTime = @WorkStart
END
IF (@FinishTime>@WorkFinish)
BEGIN
SET @FinishTime=@WorkFinish
END
DECLARE @CurrentDate DATE
SET @CurrentDate = CONVERT(DATE, @StartDate, 112)
DECLARE @LastDate DATE
SET @LastDate = CONVERT(DATE, @FinishDate, 112)
WHILE(@CurrentDate<=@LastDate)
BEGIN
IF (DATEPART(dw, @CurrentDate)!=1 AND DATEPART(dw, @CurrentDate)!=7)
BEGIN
IF (@CurrentDate!=@FirstDay) AND (@CurrentDate!=@LastDay)
BEGIN
SET @Temp = (@Temp + (9*60))
END
--IF it starts at startdate and it finishes not this date find diff between work finish and start as minutes
ELSE IF (@CurrentDate=@FirstDay) AND (@CurrentDate!=@LastDay)
BEGIN
SET @Temp = @Temp + DATEDIFF(MINUTE, @StartTime, @WorkFinish)
END
ELSE IF (@CurrentDate!=@FirstDay) AND (@CurrentDate=@LastDay)
BEGIN
SET @Temp = @Temp + DATEDIFF(MINUTE, @WorkStart, @FinishTime)
END
--IF it starts and finishes in the same date
ELSE IF (@CurrentDate=@FirstDay) AND (@CurrentDate=@LastDay)
BEGIN
SET @Temp = DATEDIFF(MINUTE, @StartDate, @FinishDate)
END
END
SET @CurrentDate = DATEADD(day, 1, @CurrentDate)
END
-- Return the result of the function
IF @Temp<0
BEGIN
SET @Temp=0
END
RETURN @Temp
END
GO
答案 6 :(得分：2)
ALTER FUNCTION WorkTime_fn (@StartDate DATETIME, @FinishDate DATETIME)
RETURNS VARCHAR(9)
AS
BEGIN
DECLARE @Temp BIGINT
SET @Temp=0
DECLARE @FirstDay VARCHAR(9)
SET @FirstDay = CONVERT(VARCHAR(9),@StartDate, 112)
DECLARE @LastDay VARCHAR(9)
SET @LastDay = CONVERT(VARCHAR(9),@FinishDate, 112)
DECLARE @StartTime VARCHAR(9)
SET @StartTime = CONVERT(VARCHAR(9),@StartDate, 108)
DECLARE @FinishTime VARCHAR(9)
SET @FinishTime = CONVERT(VARCHAR(9),@FinishDate, 108)
DECLARE @WorkStart VARCHAR(9)
SET @WorkStart = '09:30:00'
DECLARE @WorkFinish VARCHAR(9)
SET @WorkFinish = '17:30:00'
IF (@StartTime<@WorkStart)
BEGIN
SET @StartTime = @WorkStart
END
IF (@FinishTime>@WorkFinish)
BEGIN
SET @FinishTime=@WorkFinish
END
DECLARE @CurrentDate VARCHAR(9)
SET @CurrentDate = CONVERT(VARCHAR(9),@StartDate, 112)
DECLARE @LastDate VARCHAR(9)
SET @LastDate = CONVERT(VARCHAR(9),@FinishDate, 112)
WHILE(@CurrentDate<=@LastDate)
BEGIN
IF (DATEPART(dw, @CurrentDate)!=1 AND DATEPART(dw, @CurrentDate)!=7)
BEGIN
IF (@CurrentDate!=@FirstDay) AND (@CurrentDate!=@LastDay)
BEGIN
SET @Temp = (@Temp + (8*60))
END
ELSE IF (@CurrentDate=@FirstDay) AND (@CurrentDate!=@LastDay)
BEGIN
SET @Temp = @Temp + DATEDIFF(MINUTE, @StartTime, @WorkFinish)
END
ELSE IF (@CurrentDate!=@FirstDay) AND (@CurrentDate=@LastDay)
BEGIN
SET @Temp = @Temp + DATEDIFF(MINUTE, @WorkStart, @FinishTime)
END
ELSE IF (@CurrentDate=@FirstDay) AND (@CurrentDate=@LastDay)
BEGIN
SET @Temp = DATEDIFF(MINUTE, @StartTime, @FinishTime)
END
END
SET @CurrentDate = CONVERT(VARCHAR(9),DATEADD(day, 1, @CurrentDate),112)
END
Return @TEMP
END
答案 7 :(得分：1)
这是一个内联版本 Start / EndDateTime如2015-03-16 09：52：24.000 开始/结束时间（营业时间），如07:00:00 它很笨重但可以在你的选择声明中使用
我也会在功能版中发布它。
Case when <StartDate>= <EndDate> then 0
When Convert(date,<StartDate>) = Convert(date,<EndDate>) Then
IIF( DATEPART(Dw,<StartDate>) in(1,7)
or Convert(time,<StartDate>) > Convert(time,<EndTime>)
or Convert(time,<EndDate>) < Convert(time,<StartTime>),0,
DateDiff(S,IIF(Convert(time,<StartDate>) < Convert(time,<StartTime>),Convert(time,<StartTime>),Convert(time,<StartDate>))
,IIF(Convert(time,<EndDate>) > Convert(time,<EndTime>), Convert(time,<EndTime>), Convert(time,<EndDate>))))
when Convert(date,<StartDate>) <> Convert(date,<EndDate>) then
IIF(DATEPART(Dw,<StartDate>) in(1,7) or Convert(time,<StartDate>) > Convert(time,<EndTime>),0 ,DateDiff(S,IIF(Convert(time,<StartDate>) < Convert(time,<StartTime>),Convert(time,<StartTime>),Convert(time,<StartDate>)), Convert(time,<EndTime>)))
+ IIF(DATEPART(Dw,<EndDate>) in(1,7) or Convert(time,<EndDate>) < Convert(time,<StartTime>),0,DateDiff(S,Convert(time,<StartTime>),IIF(Convert(time,<EndDate>) > Convert(time,<EndTime>), Convert(time,<EndTime>), Convert(time,<EndDate>))))
else -333
end --as pday
+IIF(DatePart(wEEk,<StartDate>) = DatePart(wEEk,<EndDate>)
,0, (DateDiff(wk,dateadd(d,-datepart(dw,<StartDate>),dateadd(ww,1,<StartDate>)),DATEADD(wk, DATEDIFF(wk, 6, <EndDate>), 6)-1) * 5)) * Datediff(S, Convert(time,<StartTime>),Convert(time,<EndTime>)) --Fullweek_days
+Case When Convert(date,<StartDate>) = Convert(date,<EndDate>) then 0
When DatePart(wEEk,<StartDate>) <> DatePart(wEEk,<EndDate>) then
IIF( datepart(dw,<StartDate>) = 7,0,DateDIFF(DAY,<StartDate>+1,dateadd(d,-datepart(dw,<StartDate>),dateadd(ww,1,<StartDate>)))) -- beginFulldays
+IIF( datepart(dw,<EndDate>) = 1,0,DateDIFF(DAY,DATEADD(wk, DATEDIFF(wk, 6, <EndDate>), 6),<EndDate> -1)) --Endfulldays
When DatePart(wEEk,<StartDate>) = DatePart(wEEk,<EndDate>) then
DateDiff(DAY,<StartDate>+1,<EndDate> )
ELSE -333 END * Datediff(S, Convert(time,<StartTime>),Convert(time,<EndTime>))
这是功能版本：
CREATE FUNCTION [dbo].[rsf_BusinessTime]
(
@startDateTime Datetime,
@endDateTime Datetime ,
@StartTime VarChar(12),
@EndTime VarChar(12) )
RETURNS BIGINT
As
BEGIN
Declare @totalSeconds BigInt,
@SecondsInDay int,
@dayStart Time = Convert(time,@StartTime),
@dayEnd Time =Convert(time,@EndTime),
@SatAfterStart Datetime = dateadd(d,-datepart(dw,@startDateTime),dateadd(ww,1,@startDateTime)),
@Sunbeforend Datetime = DATEADD(wk, DATEDIFF(wk, 6, @endDateTime), 6)
-- This function calculates the seconds between the start and end dates provided for business hours.
-- It only returns the time between the @start and @end time (hour of day) of the work week.
-- Weekend days are removed.
-- Holidays are not considered.
Set @SecondsInDay = Datediff(S, @dayStart,@dayEnd)
Set @totalSeconds =
--first/last/sameday
Case when @startDateTime= @endDateTime then 0
When Convert(date,@startDateTime) = Convert(date,@endDateTime) Then
IIF( DATEPART(Dw,@startDateTime) in(1,7)
or Convert(time,@startDateTime) > @dayEnd
or Convert(time,@endDateTime) < @dayStart,0,
DateDiff(S,IIF(Convert(time,@startDateTime) < @dayStart,@dayStart,Convert(time,@startDateTime))
,IIF(Convert(time,@endDateTime) > @dayEnd, @dayEnd, Convert(time,@endDateTime))))
when Convert(date,@startDateTime) <> Convert(date,@endDateTime) then
IIF(DATEPART(Dw,@startDateTime) in(1,7) or Convert(time,@startDateTime) > @dayEnd,0 ,DateDiff(S,IIF(Convert(time,@startDateTime) < @dayStart,@dayStart,Convert(time,@startDateTime)), @dayEnd))
+ IIF(DATEPART(Dw,@endDateTime) in(1,7) or Convert(time,@endDateTime) < @dayStart,0,DateDiff(S,@dayStart,IIF(Convert(time,@endDateTime) > @dayEnd, @dayEnd, Convert(time,@endDateTime))))
else -333
end --as pday
+IIF(DatePart(wEEk,@startDateTime) = DatePart(wEEk,@endDateTime)
,0, (DateDiff(wk,@SatAfterStart,@Sunbeforend-1) * 5)) * @SecondsInDay --Fullweek_days
+Case When Convert(date,@startDateTime) = Convert(date,@endDateTime) then 0
When DatePart(wEEk,@startDateTime) <> DatePart(wEEk,@endDateTime) then
IIF( datepart(dw,@startDateTime) = 7,0,DateDIFF(DAY,@startDateTime+1,@SatAfterStart)) -- beginFulldays
+IIF( datepart(dw,@endDateTime) = 1,0,DateDIFF(DAY,@Sunbeforend,@endDateTime -1)) --Endfulldays
When DatePart(wEEk,@startDateTime) = DatePart(wEEk,@endDateTime) then
DateDiff(DAY,@startDateTime+1,@endDateTime )
ELSE -333 END * @SecondsInDay
Return @totalSeconds
END
答案 8 :(得分：1)
问题是不应该考虑公众假期，所以这个答案就是这样 - 计算考虑周末的营业时间，但忽略可能的公众假期。
它还假设给定的开始和结束日期/时间是在营业时间内。
通过这种假设，代码并不关心工作日开始或结束的时间，它只关心每天的工作时间总数。在您的示例中，09:00至17:00之间有8个工作小时。它不一定是一个整数。下面的公式以一分钟的精度计算它，但是使它成为一秒或任何其他精度是微不足道的。
如果您需要考虑公共假期，您需要有一个单独的表格，列出公共假期的日期，这些日期可能因年度和州与州或国家而异。主要公式可能保持不变，但您需要从公布假期的结果小时数中减去指定的日期范围。
公式
SELECT
DATEDIFF(minute, StartDT, EndDT) / 60.0
- DATEDIFF(day, StartDT, EndDT) * 16
- DATEDIFF(week, StartDT, EndDT) * 16 AS BusinessHours
FROM T
要了解它的工作原理，让我们创建一个表格，其中包含一些涵盖各种情况的样本数据：
DECLARE @T TABLE (StartDT datetime2(0), EndDT datetime2(0));
INSERT INTO @T VALUES
('2012-03-05 09:00:00', '2012-03-05 15:00:00'), -- simple part of the same day
('2012-03-05 10:00:00', '2012-03-06 10:00:00'), -- full day across the midnight
('2012-03-05 11:00:00', '2012-03-06 10:00:00'), -- less than a day across the midnight
('2012-03-05 10:00:00', '2012-03-06 15:00:00'), -- more than a day across the midnight
('2012-03-09 16:00:00', '2012-03-12 10:00:00'), -- over the weekend, less than 7 days
('2012-03-06 16:00:00', '2012-03-15 10:00:00'), -- over the weekend, more than 7 days
('2012-03-09 16:00:00', '2012-03-19 10:00:00'); -- over two weekends
查询
SELECT
StartDT,
EndDT,
DATEDIFF(minute, StartDT, EndDT) / 60.0
- DATEDIFF(day, StartDT, EndDT) * 16
- DATEDIFF(week, StartDT, EndDT) * 16 AS BusinessHours
FROM @T;
产生以下结果：
+---------------------+---------------------+---------------+
| StartDT | EndDT | BusinessHours |
+---------------------+---------------------+---------------+
| 2012-03-05 09:00:00 | 2012-03-05 15:00:00 | 6.000000 |
| 2012-03-05 10:00:00 | 2012-03-06 10:00:00 | 8.000000 |
| 2012-03-05 11:00:00 | 2012-03-06 10:00:00 | 7.000000 |
| 2012-03-05 10:00:00 | 2012-03-06 15:00:00 | 13.000000 |
| 2012-03-09 16:00:00 | 2012-03-12 10:00:00 | 2.000000 |
| 2012-03-06 16:00:00 | 2012-03-15 10:00:00 | 50.000000 |
| 2012-03-09 16:00:00 | 2012-03-19 10:00:00 | 42.000000 |
+---------------------+---------------------+---------------+
它有效，因为在SQL Server DATEDIFF中返回指定的 startdate 和 enddate datepart 边界的计数>
每天有8个工作小时。我计算两个日期之间的总小时数，然后减去中午的数量乘以每天16个非营业时间，然后减去周末的数量乘以16（周六+周日的8 + 8个工作小时）。
答案 9 :(得分：0)
我之前实际上已经做过这个，考虑到所有变量（周末，假期等）的营业时间非常困难，我认为这个任务最好在SQL之外完成
答案 10 :(得分：0)
另一种思维方式，如果您的星期一是星期一，则以下功能正常工作，否则您应该将相关的行（包括（6,7））更改为当地周末的日子
create function fn_worktime(@Datetime1 DateTime,@Datetime2 DateTime)
Returns BigInt
as
Begin
Declare
@Date1 Date,
@Date2 Date,
@DateIndex Date,
@minutes int,
@lastDayMinutes int,
@StartTime int , --in minutes
@FinishTime int ,--in minutes
@WorkDayLong int --in minutes
Set @StartTime =8 * 60 + 30 -- 8:30
Set @FinishTime =17* 60 + 30 -- 17:30
Set @WorkDayLong =@FinishTime - @StartTime
Set @Date1 = Convert(Date,@DateTime1)
Set @Date2 = Convert(Date,@DateTime2)
Set @minutes=DateDiff(minute,@DateTime1,DateAdd(MINUTE,@FinishTime ,convert(DateTime,@Date1)))
if @minutes<0 OR DatePart(dw,@Date1) in (6,7) -- you can even check holdays here. '(6 Saturday,7 Sunday) according to SET DATEFIRST 1'
Set @minutes=0
Set @DateIndex=DateAdd(day,1,@Date1)
While @DateIndex<@Date2
Begin
if DatePart(dw,@DateIndex) not in (6,7) -- you can even check holdays here. '(6 Saturday,7 Sunday) according to SET DATEFIRST 1'
set @minutes=@minutes+@WorkDayLong
Set @DateIndex=DateAdd(day,1,@DateIndex)
End
if DatePart(dw,@DateIndex) not in (6,7) -- you can even check holdays here
Begin
set @lastDayMinutes=DateDiff(minute,DateAdd(MINUTE ,@StartTime ,convert(DateTime,@Date2)),@DateTime2)
if @lastDayMinutes>@WorkDayLong
set @lastDayMinutes=@WorkDayLong
if @Date1<>@Date2
set @minutes=@minutes+@lastDayMinutes
Else
Set @minutes=@minutes+@lastDayMinutes-@WorkDayLong
End
return @minutes
End
答案 11 :(得分：0)
您对此解决方案有何看法？
不使用循环＆＃34;而＆＃34;。
create function dbo.WorkingHoursBetweenDates ( @StartDate datetime, @EndDate datetime, @StartTime time, @EndTime time )
returns decimal ( 10, 2 )
as
begin
return
case
when @EndTime < @StartTime or @EndDate < @StartDate then
0
else
round
( ( dbo.WorkingDaysBetweenDates(@StartDate, @EndDate) -
( dbo.WorkingDaysBetweenDates(@StartDate, @StartDate) *
case
when cast ( @StartDate as time ) > @EndTime then
1
else
datediff
( mi,
@StartTime
, case
when @StartTime > cast ( @StartDate as time ) then
@StartTime
else
cast ( @StartDate as time )
end
) /
( datediff ( mi, @StartTime, @EndTime ) + 0.0 )
end
) -
( dbo.WorkingDaysBetweenDates(@EndDate, @EndDate) *
case
when cast ( @EndDate as time ) < @StartTime then
1
else
datediff
( mi,
case
when @EndTime < cast ( @EndDate as time ) then
@EndTime
else
cast ( @EndDate as time )
end,
@EndTime
) /
( datediff ( mi, @StartTime, @EndTime ) + 0.0 )
end
)
) *
( datediff ( mi, @StartTime, @EndTime ) / 60.0 ), 2
)
end
end
------
create function dbo.WorkingDaysBetweenDates ( @StartDate date, @EndDate date )
returns int
as
begin
return
( 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 ( 1 )
from
dbo.Tb_Holidays
where
HDate between @StartDate and @EndDate
and datename(dw, HDate) not in ( 'Sunday', 'Saturday' )
)
end
答案 12 :(得分：0)
这是一种替代解决方案，不使用功能。请注意，这取决于是否存在numbers table，其中至少填充了您要跟踪的任务可能需要的最长天数。
这不考虑公众假期。如果你不在周末工作，在@OpeningHours表变量中将开始和结束时间设置为午夜应该可以胜任。
我已经对8500行“现实世界”进行了测试。数据，发现它是高效的。
DECLARE @OpeningHours TABLE ([DayOfWeek] INTEGER, OpeningTime TIME(0), ClosingTime TIME(0));
INSERT
@OpeningHours ([DayOfWeek], OpeningTime, ClosingTime)
VALUES
(1, '10:00', '16:00') -- Sun
, (2, '06:30', '23:00') -- Mon
, (3, '06:30', '23:00') -- Tue
, (4, '06:30', '23:00') -- Wed
, (5, '06:30', '23:00') -- Thu
, (6, '06:30', '23:00') -- Fri
, (7, '08:00', '20:00'); -- Sat
DECLARE @Tasks TABLE ([Description] VARCHAR(50), CreatedDateTime DATETIME, CompletedDateTime DATETIME);
INSERT
@Tasks ([Description], CreatedDateTime, CompletedDateTime)
VALUES
('Make tea', '20170404 10:00', '20170404 10:12')
, ('Make coffee', '20170404 23:35', '20170405 06:32')
, ('Write complex SQL query', '20170406 00:00', '20170406 23:32')
, ('Rewrite complex SQL query', '20170406 23:50', '20170410 10:50');
SELECT
WorkingMinutesToRespond =
SUM(CASE WHEN CAST(Tasks.CreatedDateTime AS DATE) = CAST(Tasks.CompletedDateTime AS DATE) THEN
CASE WHEN CAST(Tasks.CreatedDateTime AS TIME) < OpeningHours.OpeningTime THEN
-- Task created before opening time
DATEDIFF(MINUTE, OpeningHours.OpeningTime, CAST(Tasks.CompletedDateTime AS TIME))
ELSE
DATEDIFF(MINUTE, Tasks.CreatedDateTime, Tasks.CompletedDateTime)
END
ELSE
CASE WHEN Tasks.CoveredDate = CAST(Tasks.CreatedDateTime AS DATE) THEN
-- This is the day the task was created
CASE WHEN CAST(Tasks.CreatedDateTime AS TIME(0)) > OpeningHours.ClosingTime THEN
0 -- after working hours
ELSE
-- during or before working hours
CASE WHEN CAST(Tasks.CreatedDateTime AS TIME(0)) < OpeningHours.OpeningTime THEN
-- before opening time; take the whole day into account
DATEDIFF(MINUTE, OpeningHours.OpeningTime, OpeningHours.ClosingTime)
ELSE
-- during opening hours; take part of the day into account
DATEDIFF(MINUTE, CAST(Tasks.CreatedDateTime AS TIME), OpeningHours.ClosingTime)
END
END
ELSE
-- This is the day the task was completed
CASE WHEN Tasks.CoveredDate = CAST(Tasks.CompletedDateTime AS DATE) THEN
CASE WHEN CAST(Tasks.CompletedDateTime AS TIME(0)) < OpeningHours.OpeningTime THEN
0 -- before working hours (unlikely to occur)
ELSE
-- during or after working hours
CASE WHEN CAST(Tasks.CompletedDateTime AS TIME(0)) > OpeningHours.ClosingTime THEN
-- after closing time (also unlikely); take the whole day into account
DATEDIFF(MINUTE, OpeningHours.OpeningTime, OpeningHours.ClosingTime)
ELSE
-- during opening hours; take part of the day into account
DATEDIFF(MINUTE, OpeningHours.OpeningTime, CAST(Tasks.CompletedDateTime AS TIME(0)))
END
END
ELSE
DATEDIFF(MINUTE, OpeningHours.OpeningTime, OpeningHours.ClosingTime)
END
END
END)
, Tasks.Description
, Tasks.CreatedDateTime
, Tasks.CompletedDateTime
FROM
(
SELECT
Tasks.Description
, Tasks.CreatedDateTime
, Tasks.CompletedDateTime
, CoveredDate = CAST(DATEADD(DAY, Numbers.Number, Tasks.CreatedDateTime) AS DATE)
FROM
@Tasks Tasks
INNER JOIN (SELECT * FROM Numbers WHERE Number >= 0) Numbers ON DATEDIFF(DAY, Tasks.CreatedDateTime, Tasks.CompletedDateTime) >= Numbers.Number
) Tasks
INNER JOIN @OpeningHours OpeningHours ON DATEPART(WEEKDAY, Tasks.CoveredDate) = OpeningHours.[DayOfWeek]
GROUP BY
Tasks.Description
, Tasks.CreatedDateTime
, Tasks.CompletedDateTime
ORDER BY
Tasks.CompletedDateTime;
答案 13 :(得分：0)
我今天写的考虑银行假期的版本。注意：此方法尚未经过全面测试，无疑会得到改进。
CREATE FUNCTION [dbo].[WorkingHoursBetween2Dates]
(
@dtFrom datetime,
@dtTo datetime
)
RETURNS INT
BEGIN
DECLARE @tblDates AS TABLE (DateValue DATE)
DECLARE @dFrom date = @dtFrom
DECLARE @dTo date = @dtTo
DECLARE @intDays int
DECLARE @intHours int = 0
DECLARE @dFromWorkday bit = CASE WHEN (DATENAME(WEEKDAY, @dFrom) IN ('Saturday','Sunday')) OR EXISTS (SELECT * FROM dbo.BankHolidays WHERE BankHolidayDate = @dFrom) THEN 0 ELSE 1 END
DECLARE @dToWorkday bit = CASE WHEN (DATENAME(WEEKDAY, @dTo) IN ('Saturday','Sunday')) OR EXISTS (SELECT * FROM dbo.BankHolidays WHERE BankHolidayDate = @dTo) THEN 0 ELSE 1 END
IF DATEPART(HOUR,@dtFrom) < 9
SET @dtFrom = DATEADD(HOUR,9,CAST(CAST(@dtFrom AS DATE) AS DATETIME))
ELSE
IF DATEPART(HOUR,@dtFrom) > 17
SET @dtFrom = DATEADD(HOUR,17,CAST(CAST(@dtFrom AS DATE) AS DATETIME))
IF DATEPART(HOUR,@dtTo) < 9
SET @dtTo = DATEADD(HOUR,9,CAST(CAST(@dtTo AS DATE) AS DATETIME))
ELSE
IF DATEPART(HOUR,@dtTo) > 17
SET @dtTo = DATEADD(HOUR,17,CAST(CAST(@dtTo AS DATE) AS DATETIME))
WHILE @dFrom <= @dTo
BEGIN
INSERT INTO @tblDates
(
DateValue
)
SELECT @dFrom
WHERE NOT ((DATENAME(WEEKDAY, @dFrom) IN ('Saturday','Sunday')) OR EXISTS (SELECT * FROM dbo.BankHolidays WHERE BankHolidayDate = @dFrom))
SET @dFrom = DATEADD(DAY,1,@dFrom)
END
SET @intDays = CASE WHEN EXISTS(SELECT * FROM @tblDates) THEN (SELECT COUNT(*) FROM @tblDates) - 1 ELSE 0 END
IF @intDays = 0
BEGIN
IF @dFromWorkday = 1
IF DATEPART(HOUR,@dtFrom) < 17
BEGIN
IF DATEDIFF(DAY,@dtFrom,@dtTo)=0
SET @intHours = DATEDIFF(HOUR,@dtFrom,@dtTo)
ELSE
SET @intHours = DATEDIFF(HOUR,@dtFrom,DATEADD(HOUR,17,CAST(CAST(@dtFrom AS DATE) AS DATETIME)))
END
IF @dToWorkday = 1 AND DATEDIFF(DAY,@dtFrom,@dtTo)<>0
IF DATEPART(HOUR,@dtTo) >= 17
SET @intHours = @intHours + 8
ELSE
IF DATEPART(HOUR,@dtTo) > 9
SET @intHours = @intHours + DATEPART(HOUR,@dtTo) - 9
END
ELSE
BEGIN
IF @dFromWorkday = 1
IF DATEPART(HOUR,@dtFrom) < 17
BEGIN
SET @intHours = DATEDIFF(HOUR,@dtFrom,DATEADD(HOUR,17,CAST(CAST(@dtFrom AS DATE) AS DATETIME)))
SET @intDays = @intDays - 1
END
IF DATEPART(HOUR,@dtTo) < 17
SET @intHours = @intHours + (@intDays * 8) + CASE WHEN @dToWorkday = 1 THEN DATEDIFF(HOUR,DATEADD(HOUR,9,CAST(CAST(@dtTo AS DATE) AS DATETIME)), @dtTo) ELSE 0 END
ELSE
SET @intHours = @intHours + ((@intDays + 1) * 8)
END
RETURN (@intHours)
END
答案 14 :(得分：0)
我刚刚遇到了这个问题，并提出了以下解决方案。它将作为工作日的开始和结束时间，计算出部分工作日使用的分钟数，如果工作时间超过 x 小时，则删除午休时间，并删除周末和银行假期。
这是锻炼： 日期从“2020-09-09 07:49”到“2020-12-18 08:27:00.000” = 34598 分钟
<块引用>= 101 days
- 28 weekends
= 73 days (71 minus start and end partial days)
= 34080 minutes (568 hours x 71 whole days @ 8 hours per day)
+ start date 07:49 to 17:00 with an hour for lunch (8 hours 11 minutes) = 491 (551 minutes - 60 for lunch)
+ end date 08:00 to 08:27 = 27 minutes
= 34080 + 491 + 27 = 34598
CREATE TABLE bankHoliday (
bhDate DATE
) ON [PRIMARY]
-- 2021
INSERT INTO bankHoliday (bhDate) VALUES ('2021-01-01' );
INSERT INTO bankHoliday (bhDate) VALUES ('2021-04-02' );
INSERT INTO bankHoliday (bhDate) VALUES ('2021-04-05' );
INSERT INTO bankHoliday (bhDate) VALUES ('2021-05-03' );
INSERT INTO bankHoliday (bhDate) VALUES ('2021-05-31' );
INSERT INTO bankHoliday (bhDate) VALUES ('2021-08-30' );
INSERT INTO bankHoliday (bhDate) VALUES ('2021-12-27' );
INSERT INTO bankHoliday (bhDate) VALUES ('2021-12-28' );
DROP FUNCTION dbo.[fn_GetTotalWorkingMinutes];
GO
CREATE FUNCTION [fn_GetTotalWorkingMinutes]
(
@DateFrom Datetime,
@DateTo Datetime
)
RETURNS INT
AS
BEGIN
DECLARE @StartOfDay TIME = '08:00:00'
DECLARE @EndOfDay TIME = '17:00:00'
DECLARE @Lunch INT = 60
DECLARE @ThresholdForLunch INT = 300 -- 5 hours x 60 mins, assume x minutes for lunch (working 8:00AM to after 1:00PM, or before 12:00PM to 5:00pm will probably include lunch)
DECLARE @WorkingMinutes INT = 0
;WITH seq(n) AS
(
-- Get an integer sequence number starting from zero, per day between From and To dates
SELECT 0 UNION ALL SELECT n + 1 FROM seq
WHERE n < DATEDIFF(DAY, @DateFrom, @DateTo)
),
d(Id,calDate,dayNum) AS
(
-- Return the sequence as an Id, the dates between From and To, and the Day-of-week number (1 = Sunday, 7 = Saturday)
SELECT n AS Id
, DATEADD(DAY, n, @DateFrom) AS calDate
, DATEPART(dw, DATEADD( DAY, n, @DateFrom)) AS dayNum
FROM seq
),
src AS
(
SELECT Id
, CAST(CONVERT(NVARCHAR(10), d.calDate, 126) AS date) AS calDate
, dayNum
, CASE
WHEN FORMAT(d.calDate, 'yyyy-MM-dd' ) = FORMAT(@DateFrom, 'yyyy-MM-dd' ) THEN @DateFrom -- Return the Date From with Start Time
ELSE CAST(CONVERT(NVARCHAR(10), d.calDate, 126) AS datetime) + CAST(@StartOfDay AS datetime) -- Add the Start of Day Time to each day
END AS calStartDateTime
, CASE
WHEN FORMAT(d.calDate, 'yyyy-MM-dd' ) = FORMAT(@DateTo, 'yyyy-MM-dd' ) THEN @DateTo -- Return the Date To with End Time
ELSE CAST(CONVERT(NVARCHAR(10), d.calDate, 126) AS datetime) + CAST(@EndOfDay AS datetime) -- Add the End of Day Time to each day
END AS calEndDateTime
FROM d
-- Exclude Bank Holidays
LEFT JOIN dbo.bankHoliday bh
ON bh.bhDate = d.caldate
WHERE bh.bhDate IS NULL -- Exclude Bank Holidays
AND d.dayNum > 1 -- After Sunday
AND d.dayNum < 7 -- Before Saturday
)
/* For Debugging
SELECT Id
, calDate
, calStartDateTime
, calEndDateTime
, CASE
WHEN DATEDIFF(minute, calStartDateTime, calEndDateTime) > 5 THEN DATEDIFF(minute, calStartDateTime, calEndDateTime) - @Lunch -- Take an hour off for lunch if more than 5 hours have been worked
ELSE DATEDIFF(minute, calStartDateTime, calEndDateTime)
END AS WorkingHours
*/
SELECT @WorkingMinutes = SUM(
CASE
WHEN DATEDIFF(minute, calStartDateTime, calEndDateTime) > @ThresholdForLunch THEN DATEDIFF(minute, calStartDateTime, calEndDateTime) - @Lunch -- Take an hour off for lunch if more than 5 hours have been worked
ELSE DATEDIFF(minute, calStartDateTime, calEndDateTime)
END
)
FROM src
--ORDER BY calDate
OPTION (MAXRECURSION 0);
RETURN (@workingMinutes);
END
GO
SELECT requested
, solved
, dbo.[fn_GetTotalWorkingMinutes](requested, solved)
FROM MyTable