我需要计算两个日期之间的 DateDiff（小时），但仅限于工作时间（8:30 - 16:00，无周末）。然后根据下面的示例将此结果放入 Reaction_Time 列。
ID 日期 Reaction_Time 过期 1 2003 年 4 月 29 日 15:00:00 1 2003 年 4 月 30 日 11:00:00 3:30 2 2003 年 4 月 30 日 14:00:00 2 01.05.2003 14:00:00 7:30 是
*注意：我没有检查示例中的日期是否为假期。
我正在使用 SQL Server 2005
这将与一个更大的查询相结合，但现在我只需要这个就可以开始了，我将尝试弄清楚如何自己将它们组合在一起。谢谢您的帮助！
编辑：嘿，谢谢大家的回复。但由于 SQL 端解决方案的明显复杂性，我们决定在 Excel 中执行此操作，因为无论如何报表都将移动到该位置。很抱歉给您带来麻烦，但我真的认为它会比这更简单。事实上，我们只是没有时间。
我建议构建一个用户定义的函数，根据您的规则计算营业时间的日期差异。
SELECT
Id,
MIN(Date) DateStarted,
MAX(Date) DateCompleted,
dbo.udfDateDiffBusinessHours(MIN(Date), MAX(Date)) ReactionTime
FROM
Incident
GROUP BY
Id
我不确定你的Overdue价值来自哪里，所以我在我的例子中把它省略了。
在函数中，您可以编写比在查询中更具表现力的 SQL，并且不会用业务规则阻塞查询，从而使其难以维护。
一个函数也可以很容易地被重用。扩展它以包括对假期的支持（我在Holidays这里考虑一张桌子）不会太难。无需更改难以阅读的嵌套 SELECT/CASE WHEN 构造，就可以进行进一步的改进，这将是替代方案。
如果我今天有时间，我会考虑编写一个示例函数。
编辑：这里有一些花里胡哨的东西，在周末透明地计算：
ALTER FUNCTION dbo.udfDateDiffBusinessHours (
@date1 DATETIME,
@date2 DATETIME
) RETURNS DATETIME AS
BEGIN
DECLARE @sat INT
DECLARE @sun INT
DECLARE @workday_s INT
DECLARE @workday_e INT
DECLARE @basedate1 DATETIME
DECLARE @basedate2 DATETIME
DECLARE @calcdate1 DATETIME
DECLARE @calcdate2 DATETIME
DECLARE @cworkdays INT
DECLARE @cweekends INT
DECLARE @returnval INT
SET @workday_s = 510 -- work day start: 8.5 hours
SET @workday_e = 960 -- work day end: 16.0 hours
-- calculate Saturday and Sunday dependent on SET DATEFIRST option
SET @sat = CASE @@DATEFIRST WHEN 7 THEN 7 ELSE 7 - @@DATEFIRST END
SET @sun = CASE @@DATEFIRST WHEN 7 THEN 1 ELSE @sat + 1 END
SET @calcdate1 = @date1
SET @calcdate2 = @date2
-- @date1: assume next day if start was after end of workday
SET @basedate1 = DATEADD(dd, 0, DATEDIFF(dd, 0, @calcdate1))
SET @calcdate1 = CASE WHEN DATEDIFF(mi, @basedate1, @calcdate1) > @workday_e
THEN @basedate1 + 1
ELSE @calcdate1
END
-- @date1: if Saturday or Sunday, make it next Monday
SET @basedate1 = DATEADD(dd, 0, DATEDIFF(dd, 0, @calcdate1))
SET @calcdate1 = CASE DATEPART(dw, @basedate1)
WHEN @sat THEN @basedate1 + 2
WHEN @sun THEN @basedate1 + 1
ELSE @calcdate1
END
-- @date1: assume @workday_s as the minimum start time
SET @basedate1 = DATEADD(dd, 0, DATEDIFF(dd, 0, @calcdate1))
SET @calcdate1 = CASE WHEN DATEDIFF(mi, @basedate1, @calcdate1) < @workday_s
THEN DATEADD(mi, @workday_s, @basedate1)
ELSE @calcdate1
END
-- @date2: assume previous day if end was before start of workday
SET @basedate2 = DATEADD(dd, 0, DATEDIFF(dd, 0, @calcdate2))
SET @calcdate2 = CASE WHEN DATEDIFF(mi, @basedate2, @calcdate2) < @workday_s
THEN @basedate2 - 1
ELSE @calcdate2
END
-- @date2: if Saturday or Sunday, make it previous Friday
SET @basedate2 = DATEADD(dd, 0, DATEDIFF(dd, 0, @calcdate2))
SET @calcdate2 = CASE DATEPART(dw, @calcdate2)
WHEN @sat THEN @basedate2 - 0.00001
WHEN @sun THEN @basedate2 - 1.00001
ELSE @date2
END
-- @date2: assume @workday_e as the maximum end time
SET @basedate2 = DATEADD(dd, 0, DATEDIFF(dd, 0, @calcdate2))
SET @calcdate2 = CASE WHEN DATEDIFF(mi, @basedate2, @calcdate2) > @workday_e
THEN DATEADD(mi, @workday_e, @basedate2)
ELSE @calcdate2
END
-- count full work days (subtract Saturdays and Sundays)
SET @cworkdays = DATEDIFF(dd, @basedate1, @basedate2)
SET @cweekends = @cworkdays / 7
SET @cworkdays = @cworkdays - @cweekends * 2
-- calculate effective duration in minutes
SET @returnval = @cworkdays * (@workday_e - @workday_s)
+ @workday_e - DATEDIFF(mi, @basedate1, @calcdate1)
+ DATEDIFF(mi, @basedate2, @calcdate2) - @workday_e
-- return duration as an offset in minutes from date 0
RETURN DATEADD(mi, @returnval, 0)
END
该函数返回一个DATETIME值，表示从日期 0（即"1900-01-01 00:00:00"）开始的偏移量。因此，例如 8:00 小时的时间跨度是"1900-01-01 08:00:00"25 小时"1900-01-02 01:00:00"。函数结果是两个日期之间的营业时间的时差。加班无特殊处理/支持。
SELECT dbo.udfDateDiffBusinessHours('2003-04-29 15:00:00', '2003-04-30 11:00:00')
--> 1900-01-01 03:30:00.000
SELECT dbo.udfDateDiffBusinessHours('2003-04-30 14:00:00', '2003-05-01 14:00:00')
--> 1900-01-01 07:30:00.000
该函数假定下一个可用工作日 (08:30 h) 的开始时间是非工作@date1时间，而上一个可用工作日 (16:00 h) 的结束时间@date2是非工作时间。
“下一个/上一个可用”是指：
@date1是'2009-02-06 07:00:00'(Fri)，它将变成'2009-02-06 08:30:00'(Fri)@date1是'2009-02-06 19:00:00'（星期五），它将变成'2009-02-09 08:30:00'（星期一）@date2是'2009-02-09 07:00:00'(Mon)，它将变成'2009-02-06 16:00:00'(Fri)@date2是'2009-02-09 19:00:00'（星期一），它将变成'2009-02-09 16:00:00'（星期一）DECLARE @BusHourStart DATETIME, @BusHourEnd DATETIME
SELECT @BusHourStart = '08:30:00', @BusHourEnd = '16:00:00'
DECLARE @BusMinutesStart INT, @BusMinutesEnd INT
SELECT @BusMinutesStart = DATEPART(minute,@BusHourStart)+DATEPART(hour,@BusHourStart)*60,
@BusMinutesEnd = DATEPART(minute,@BusHourEnd)+DATEPART(hour,@BusHourEnd)*60
DECLARE @Dates2 TABLE (ID INT, DateStart DATETIME, DateEnd DATETIME)
INSERT INTO @Dates2
SELECT 1, '15:00:00 04/29/2003', '11:00:00 04/30/2003' UNION
SELECT 2, '14:00:00 04/30/2003', '14:00:00 05/01/2003' UNION
SELECT 3, '14:00:00 05/02/2003', '14:00:00 05/06/2003' UNION
SELECT 4, '14:00:00 05/02/2003', '14:00:00 05/04/2003' UNION
SELECT 5, '07:00:00 05/02/2003', '14:00:00 05/02/2003' UNION
SELECT 6, '14:00:00 05/02/2003', '23:00:00 05/02/2003' UNION
SELECT 7, '07:00:00 05/02/2003', '08:00:00 05/02/2003' UNION
SELECT 8, '22:00:00 05/02/2003', '23:00:00 05/03/2003' UNION
SELECT 9, '08:00:00 05/03/2003', '23:00:00 05/04/2003' UNION
SELECT 10, '07:00:00 05/02/2003', '23:00:00 05/02/2003'
-- SET DATEFIRST to U.S. English default value of 7.
SET DATEFIRST 7
SELECT ID, DateStart, DateEnd, CONVERT(VARCHAR, Minutes/60) +':'+ CONVERT(VARCHAR, Minutes % 60) AS ReactionTime
FROM (
SELECT ID, DateStart, DateEnd, Overtime,
CASE
WHEN DayDiff = 0 THEN
CASE
WHEN (MinutesEnd - MinutesStart - Overtime) > 0 THEN (MinutesEnd - MinutesStart - Overtime)
ELSE 0
END
WHEN DayDiff > 0 THEN
CASE
WHEN (StartPart + EndPart - Overtime) > 0 THEN (StartPart + EndPart - Overtime)
ELSE 0
END + DayPart
ELSE 0
END AS Minutes
FROM(
SELECT ID, DateStart, DateEnd, DayDiff, MinutesStart, MinutesEnd,
CASE WHEN(@BusMinutesStart - MinutesStart) > 0 THEN (@BusMinutesStart - MinutesStart) ELSE 0 END +
CASE WHEN(MinutesEnd - @BusMinutesEnd) > 0 THEN (MinutesEnd - @BusMinutesEnd) ELSE 0 END AS Overtime,
CASE WHEN(@BusMinutesEnd - MinutesStart) > 0 THEN (@BusMinutesEnd - MinutesStart) ELSE 0 END AS StartPart,
CASE WHEN(MinutesEnd - @BusMinutesStart) > 0 THEN (MinutesEnd - @BusMinutesStart) ELSE 0 END AS EndPart,
CASE WHEN DayDiff > 1 THEN (@BusMinutesEnd - @BusMinutesStart)*(DayDiff - 1) ELSE 0 END AS DayPart
FROM (
SELECT DATEDIFF(d,DateStart, DateEnd) AS DayDiff, ID, DateStart, DateEnd,
DATEPART(minute,DateStart)+DATEPART(hour,DateStart)*60 AS MinutesStart,
DATEPART(minute,DateEnd)+DATEPART(hour,DateEnd)*60 AS MinutesEnd
FROM (
SELECT ID,
CASE
WHEN DATEPART(dw, DateStart) = 7
THEN DATEADD(SECOND, 1, DATEADD(DAY, DATEDIFF(DAY, 0, DateStart), 2))
WHEN DATEPART(dw, DateStart) = 1
THEN DATEADD(SECOND, 1, DATEADD(DAY, DATEDIFF(DAY, 0, DateStart), 1))
ELSE DateStart END AS DateStart,
CASE
WHEN DATEPART(dw, DateEnd) = 7
THEN DATEADD(SECOND, -1, DATEADD(DAY, DATEDIFF(DAY, 0, DateEnd), 0))
WHEN DATEPART(dw, DateEnd) = 1
THEN DATEADD(SECOND, -1, DATEADD(DAY, DATEDIFF(DAY, 0, DateEnd), -1))
ELSE DateEnd END AS DateEnd FROM @Dates2
)Weekends
)InMinutes
)Overtime
)Calculation
select datediff(hh,@date1,@date2) - 16.5*(datediff(dd,@date1,@date2))
唯一的问题是它会给你 3:30 作为 3.5 小时，但你可以很容易地解决这个问题。
使用此代码：找出日期之间的周末
(
DATEDIFF(dd, open_date, zassignment_date) + 1
- ( (DATEDIFF(dd, open_date, zassignment_date) + 1)
-(DATEDIFF(wk, open_date, zassignment_date) * 2)
-(CASE WHEN DATENAME(dw, open_date) = 'Sunday' THEN 1 ELSE 0 END)
-(CASE WHEN DATENAME(dw, zassignment_date) = 'Saturday' THEN 1 ELSE 0 END) )) wk_end
假设您有工作日（及其小时数）的参考表，那么我将使用 3 阶段方法（伪 sql）
（首先排除“一天之内”的琐碎示例，因为这简化了逻辑）
-- days that are neither the start nor end (full days)
SELECT @FullDayHours = SUM(day start to day end)
FROM reference-calendar
WHERE Start >= midnight-after-start and End <= midnight-before-end
-- time after the [query start] to the end of the first working day
SELECT @FirstDayHours = [query start] to day end
FROM reference-calandar
WHERE start day
-- time from the start of the last working day to the [query end]
SELECT @LastDayHours = day start to [query end]
FROM reference-calendar
WHERE end-day
IF @FirstDayHours < 0 SET @FirstDayHours = 0 -- starts outside working time
IF @LastDayHours < 0 SET @LastDayHours = 0 -- ends outside working time
PRINT @FirstDayHours + @FullDayHours + @LastDayHours
显然，如果没有更多上下文，要正确地做有点困难......
此功能将为您提供两个给定时间之间的营业时间差异。这将根据日期部分参数返回以分钟或小时为单位的差异。
CREATE FUNCTION [dbo].[fnBusinessHoursDateDiff] (@StartTime SmallDatetime, @EndTime SmallDateTime, @DatePart varchar(2)) RETURNS DECIMAL (10,2)
AS
BEGIN
DECLARE @Minutes bigint
, @FinalNumber Decimal(10,2)
-- // Create Minute By minute table for CTE
-- ===========================================================
;WITH cteInputHours (StartTime, EndTime, NextTime) AS (
SELECT @StartTime
, @EndTime
, dateadd(mi, 1, @StartTime)
),
cteBusinessMinutes (TimeOfDay, [isBusHour], NextTime) AS(
SELECT StartTime [TimeOfDay]
, case when datepart(dw, StartTime) between 2 and 6 and convert(time,StartTime) between '08:30' and '15:59' then 1 else 0 end [isBusHour]
, dateadd(mi, 1, @StartTime) [NextTime]
FROM cteInputHours
UNION ALL
SELECT dateadd(mi, 1, (a.TimeOfDay)) [TimeOfDay]
, case when datepart(dw, a.TimeOfDay) between 2 and 6 and convert(time,dateadd(mi, 1, (a.TimeOfDay)) ) between '08:30' and '15:59' then 1 else 0 end [isBusHour]
, dateadd(mi, 2, (a.TimeOfDay)) NextTime
FROM cteBusinessMinutes a
WHERE dateadd(mi, 1, (a.TimeOfDay)) < @EndTime
)
SELECT @Minutes = count(*)
FROM cteBusinessMinutes
WHERE isBusHour = 1
OPTION (MAXRECURSION 0);
-- // Final Select
-- ===========================================================
SELECT @FinalNumber = @Minutes / (case when @DatePart = 'hh' then 60.00 else 1 end)
RETURN @FinalNumber
END