Tsql DATEDIFF仅限工作时间和天数

我试图写一份报告，但有点卡住了：/我试图显示两个日期之间的小时和分钟，但减去非工作时间

例如，一家公司在工作日的08:00到17:00之间工作，一个电话在今天的16:00被记录，明天的16:00结束，这将是24小时减去工作时间，因此将在9小时工作

我还创建了一个单独的表格，其中包含一年中除周末、工作日开始和工作日结束之外的所有日期。但我仍然要找出没有非营业时间的时间间隔

示例数据：

Call_Initiated - Call_Ended
10/05/2013 15:00 - 13/05/2013 13:00

结果我想要

Call_Initiated - Call_Ended - Time_To_Resolve
10/05/2013 15:00 - 13/05/2013 13:00 - 07

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我只是好奇你的问题，然后做了这个

也许不是最好的剧本，但它可能会给你一些解决问题的想法

它功能齐全，但我生成了日期，您可能希望使用day表

declare @callLogStart datetime = '2013-01-04 16:00'
declare @callLogEnd datetime = '2013-01-08 09:00'

;with dates(startDate, endDate)
as
(
select  cast('2013-01-01 08:00' as datetime)
        ,cast('2013-01-01 17:00' as datetime)
union all
select  DATEADD(day,1, startDate)
        ,DATEADD(day, 1, endDate)
from    dates
where   startDate < '2013-02-01 08:00'
)
,startDay
as
(
    select  *
            ,Datediff(hour, d.startDate, d.endDate) - DATEDIFF(hour, startDate, @callLogStart) as spent
    from    dates d
    where   @callLogStart between d.startDate and d.endDate
)
,endDay
as
(
    select  *
            ,Datediff(hour, d.startDate, d.endDate) - datediff(hour, @callLogEnd, endDate) as spent
    from    dates d
    where   @callLogEnd between d.startDate and d.endDate
)

select  --SUM(spent) as actualTime
        spent
        ,startDate
        ,endDate
        ,mark
from
(
    select  startDate
            ,endDate
            ,spent
            ,'start' as mark 
    from    startDay
    union
    select  startDate
            ,endDate
            ,spent
            ,'end'
    from    endDay
    union
    select  s.startDate
            ,s.endDate
            ,-Datediff(hour, s.startDate, s.endDate)
            ,'remove'
    from    startDay s
    join    endDay e
        on  s.startDate = e.startDate
        and s.endDate = e.endDate
    union
    select  startDate
            ,endDate
            ,Datediff(hour, startDate, endDate)
            ,'between'
    from    dates
    where   @callLogStart < startDate
    except
    select  startDate
            ,endDate
            ,Datediff(hour, startDate, endDate)
            ,'between'
    from    dates
    where   @callLogEnd < endDate
) x
order by    
    case mark 
        when 'start' then 0 
        when 'between' then 1 
        when 'end' then 2 
        when 'remove' then 3 
    end

---

希望能有所帮助

这稍微简单一点。只有一个select语句。我把每一步分解成它自己的一列，这样你就可以看到它是如何工作的。不过，您只需要最后一列就可以计算出小时数。它依赖于区域设置，因为它使用DateName，但只要知道DATEFIRST设置为什么，您可以在一周中的某一天将其翻转过来

此外，这不包括假期。您必须创建自己的假日表。我引用了你可以链接到最终公式的地方

只需将开始和结束日期设置为您想要使用的任何日期，然后在代码中使用它，执行查找/替换，并用字段名替换这些参数。如果您使用的是SQL Server 2008或更高版本，则可以通过将打开/关闭时间转换为时间数据类型来简化很多操作。希望这有帮助

declare @startDate datetime = '2013-09-05 10:45:00.000',
        @endDate datetime = '2013-09-06 08:15:00.000',
        @zeroDate datetime = '1900-01-01 00:00:00.000',
        @businessOpen datetime = '1900-01-01 08:00:00.000',
        @businessClose datetime = '1900-01-01 17:00:00.000',
        @hoursOpen int;

select @hoursOpen = datediff(hour, @businessOpen, @businessClose);

select @hoursOpen as hoursOpen
        , @endDate - @startDate as actualTimeCallOpen
        , datediff(week, @startDate, @endDate) as wholeWeekendsCallOpen
        , datediff(day, @startDate, @endDate) as daysCallOpen
        , (DATEDIFF(dd, @StartDate, @EndDate)) --get days apart
            -(DATEDIFF(wk, @StartDate, @EndDate) * 2) --subtract whole weekends from the date (*2 is for 2 days per weekend)
              +(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN 1 ELSE 0 END) --subtract the start date if it started on sunday (thus, partial weekend)
              -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END) --subtract the end date if it ends on saturday (again, partial weekend)
            as MthruFDaysOpen
        , datediff(hour, @startDate, @endDate) as timeHoursCallOpen
        , datediff(minute, @businessOpen, convert(datetime, '1900-01-01 ' + convert(varchar(8),@startDate,108))) / 60.0 as hoursOpenBeforeCall
        , datediff(minute, convert(datetime, '1900-01-01 ' + convert(varchar(8), @endDate, 108)), @businessClose) / 60.0 as hoursOpenAfterCall
        , (@hoursOpen - ((datediff(minute, convert(datetime, '1900-01-01 ' + convert(varchar(8), @endDate, 108)), @businessClose) + datediff(minute, @businessOpen, convert(datetime, '1900-01-01 ' + convert(varchar(8),@startDate,108)))) / 60.0)) as partialHourDay
        , ( ((DATEDIFF(dd, @StartDate, @EndDate)) --get days apart, 
            - (DATEDIFF(wk, @StartDate, @EndDate) * 2) --subtract whole weekends from the date
            + (CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN 1 ELSE 0 END) --subtract the start date if it started on sunday (thus, partial weekend)
            - (CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END) --subtract the end date if it ends on saturday (again, partial weekend)
            --If you have a table with holidays in it, you can subtract the count of holidays from this as well
            --test where the holiday is between startdate and end date and the holiday itself isn't a saturday or sunday
            ) * @hoursOpen) --multiply the whole days open times hours per day, giving us 
        + (@hoursOpen --start with hours open
            - ( -- then subtract the sum of hours the business was open before and after the call
                (datediff(minute, convert(datetime, '1900-01-01 ' + convert(varchar(8), @endDate, 108)), @businessClose) --calculate this different in minutes for greater accuracy
                    + datediff(minute, @businessOpen, convert(datetime, '1900-01-01 ' + convert(varchar(8),@startDate,108)))
                ) / 60.0) --divide by 60 to convert back to hours before subtracting from @hours open
            ) as businessTimeOpen

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谢谢你，约翰，我要吃一只鹅：