SQL Server-计算两次之间的小时数，并将这些小时分为三个时间段

Question

我正在使用SQL Server 2008，甚至不知道从哪里开始计算。非常感谢您能提供的任何帮助。

我有两列，StartTime和EndTime。我有三个时间范围：上午12:00-上午7:59，白天8:00到3:59（或15:39）pm和晚上4:00 pm（或16:00）到11:59 （或23:59）下午。

我想计算每个范围内的StartTime和EndTime之间的小时数。例如，对于下面的第一行，从6:00到17:00之间有11个小时，早上有2个小时，白天有8个小时，晚上有1个小时。因此，我需要四列：一列用于11个小时，一列用于早晨，白天和傍晚。

StartTime               EndTime
2017-10-25 06:00:00.000 2017-10-25 17:00:00.000
2017-10-26 05:30:00.000 2017-10-26 18:00:00.000
2017-10-30 07:00:00.000 2017-10-30 17:30:00.000
2017-11-01 06:00:00.000 2017-11-01 17:30:00.000
2017-10-06 04:00:00.000 2017-10-06 05:00:00.000
2016-04-28 04:00:00.000 2016-04-28 10:00:00.000
2017-06-30 04:00:00.000 2017-07-01 00:00:00.000
2016-01-26 06:30:00.000 2016-01-26 19:00:00.000
2017-08-15 07:00:00.000 2017-08-15 19:30:00.000
2016-01-28 07:00:00.000 2016-01-28 19:30:00.000

所需结果，使用第一个结果

NbrHours = 11   
MorningHrs  = 2
DayHrs  = 8
EveningHrs = 1

Answer 1

与此类似的事情应该可以帮到您...

drop table if exists #times;
go

create table #times(
    StartTime datetime ,
    EndTime datetime
);

insert #times values
('2017-10-25 06:00:00.000', '2017-10-25 17:00:00.000'),
('2017-10-26 05:30:00.000', '2017-10-26 18:00:00.000'),
('2017-10-30 07:00:00.000', '2017-10-30 17:30:00.000'),
('2017-11-01 06:00:00.000', '2017-11-01 17:30:00.000'),
('2017-10-06 04:00:00.000', '2017-10-06 05:00:00.000'),
('2016-04-28 04:00:00.000', '2016-04-28 10:00:00.000'),
('2017-06-30 04:00:00.000', '2017-07-01 00:00:00.000'),
('2016-01-26 06:30:00.000', '2016-01-26 19:00:00.000'),
('2017-08-15 07:00:00.000', '2017-08-15 19:30:00.000'),
('2016-01-28 07:00:00.000', '2016-01-28 19:30:00.000')
;

with
    a as (
        select  StartTime ,
                EndTime ,
                datepart(hour, StartTime) as StartHour ,
                datepart(hour, StartTime) + datediff(hour, StartTime, EndTime) as EndHour
        from    #times
    )
select  a.StartTime,
        a.EndTime ,
        StartHour ,
        EndHour ,
        EndHour - StartHour as Total ,
        case when EndHour > 8 then 8 else EndHour end - case when StartHour < 0 then 0 else StartHour end as Morning ,
        case when EndHour > 16 then 16 else EndHour end - case when StartHour < 8 then 8 else StartHour end as Afternoon ,
        case when EndHour > 24 then 24 else EndHour end - case when StartHour < 16 then 16 else StartHour end as Evening
from    a
;