SQL DateDiff高级用法？
https://stackoverflow.com/questions/593898
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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，该结果将被放入 Reaction_Time 列中，如下例所示。
ID Date Reaction_Time Overdue 1 29.04.2003 15:00:00 1 30.04.2003 11:00:00 3:30 2 30.04.2003 14:00:00 2 01.05.2003 14:00:00 7:30 YES
*笔记：我没有检查示例中的日期是否是假期。
我正在使用 SQL Server 2005
这将与更大的查询相结合，但现在我所需要的只是这个来开始，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 Excel 中。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。
解决方案
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
其他提示
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
我不确定你在哪里 Overdue 价值来自于，所以我在例子中省略了它。
在函数中，您可以编写比在查询中更具表现力的 SQL，并且不会因业务规则而阻塞查询，从而使其难以维护。
函数也可以很容易地被重用。将其扩展为包括对假期的支持（我正在考虑 Holidays 表在这里）不会太难。可以进行进一步的细化，而无需更改难以阅读的嵌套 SELECT/CASE WHEN 结构，这将是替代方案。
如果今天有时间，我会考虑编写一个示例函数。
编辑：这是一些花里胡哨的东西，透明地计算周末：
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 值表示距日期 0 的偏移量（即 "1900-01-01 00:00:00"）。例如，时间跨度为 8:00 小时 "1900-01-01 08:00:00" 25小时就是 "1900-01-02 01:00:00". 。函数结果就是时间 业务上的差异 两个日期之间的小时数。没有对加班的特殊处理/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
该函数假设下一个可用工作日（08:30）开始，当 @date1 是下班时间，并且是前一个可用工作日结束时（16:00） @date2 是下班时间。
“下一个/上一个可用”是指：
- 如果
@date1是'2009-02-06 07:00:00'（周五），它将变成'2009-02-06 08:30:00'（周五） - 如果
@date1是'2009-02-06 19:00:00'（周五），它将变成'2009-02-09 08:30:00'（周一） - 如果
@date2是'2009-02-09 07:00:00'（周一），它将变成'2009-02-06 16:00:00'（周五） - 如果
@date2是'2009-02-09 19:00:00'（周一），它将变成'2009-02-09 16:00:00'（周一）
select datediff(hh,@date1,@date2) - 16.5*(datediff(dd,@date1,@date2))
，唯一的缺点是，它会给你3:30 3.5小时，但您可以修复容易。
使用此代码：找出周末在日期间的
(
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
假设你有的工作日（和它们的小时）的参考表，然后我会用3阶段的方法（伪SQL）
（第一排除“在一天”简单的例子中，由于简化了逻辑）
-- 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，这是一个有点难以没有更多的情况下做正确...
这个功能会给你在给定的两个时间之间的工作时间的差异。这将返回基于日期部分参数在几分钟或几小时的差。
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
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