在论坛中出现的比较难的sql问题：35(时间间隔计算问题)

最近，在论坛中，遇到了不少比较难的sql问题，虽然自己都能解决，但发现过几天后，就记不起来了，也忘记解决的方法了。

所以，觉得有必要记录下来，这样以后再次碰到这类问题，也能从中获取解答的思路。

时间间隔计算 http://bbs.csdn.net/topics/390608930 
这个问题非常复杂。

start_time                  end_time

2013-09-11 17:26:02.382      2013-09-24 10:38:01.41

2013-09-18 17:02:40.444      2013-09-22 15:27:58.984

2013-09-18 08:21:32.036      2013-09-22 15:31:52.499

2013-09-13 16:28:29.832      2013-09-16 09:41:47.988

2013-09-09 10:59:59.835      2013-09-10 14:06:21.223

要求计算这两个列的时间差 但是要去除9月份的正常休假并且只计算正常工作时间（上午8:30--12:00 下午14:00--18:00）

计算结果如下：

start_time                  end_time                        diff_time（小时）

2013-09-11 17:26:02.382      2013-09-24 10:38:01.41         55.1

2013-09-18 17:02:40.444      2013-09-22 15:27:58.984        5.9

2013-09-18 08:21:32.036      2013-09-22 15:31:52.499        12.5

2013-09-13 16:28:29.832      2013-09-16 09:41:47.988        2.7

2013-09-09 10:59:59.835      2013-09-10 14:06:21.223        1.1

请各位大大帮忙看看这个时间差应该怎么计算  谢谢了

 

我的解法： 

if object_id('tab') is not null
drop table tab

if object_id('holiday') is not null
drop table holiday

go
create table tab(start_time datetime,end_time datetime)

insert into tab
select '2013-09-11 17:26:02.382','2013-09-24 10:38:01.41' union
select '2013-09-18 17:02:40.444','2013-09-22 15:27:58.984' union
select '2013-09-18 08:21:32.036','2013-09-22 15:31:52.499' union
select '2013-09-13 16:28:29.832','2013-09-16 09:41:47.988' union
select '2013-09-09 10:59:59.835','2013-09-09 14:06:21.223'

create table holiday(h_date datetime)

insert into holiday
select '2013-09-01' union
select '2013-09-07' union
select '2013-09-08' union
select '2013-09-14'union
select '2013-09-15'union
select '2013-09-19'union
select '2013-09-20'union
select '2013-09-21'union
select '2013-09-29'
go

WITH calendar --产生日历
AS
(
SELECT CAST('2013-09-01' as varchar(10)) AS r --月份的开始日期

UNION ALL

SELECT convert(VARCHAR(10),dateadd(day,1,r),120)
FROM calendar
WHERE r < '2013-09-30' --月份的结束日期
),

tt --计算时间间隔，单位为秒
as
(
SELECT t.start_time,
t.end_time,

c.r,
h.h_date,

/*
通过tab表和calendar表的关联，就能把开始时间到结束时间，多对应的多天，
都给关联出来，
比如开始时间 2013-09-18 08:21:32.037 结束时间 2013-09-22 15:31:52.500，
其实就是，18、19、20、21、22这一共5天，会由原来的1条记录，现在变为5条记录。

如果h_date为null，说明这一天不是假日,
就需要计算时间间隔，有几种可能性：
1.开始时间和结束时间，在同一天的
2.当前日期和开始日期相同
3.当前日期和结束日期相同
4.当前日期是在开始日期和结束日期之间的某天

如果h_date是null，那么返回0，说明是节假日，就不用计算时间间隔了
*/
case when h_date IS null and
convert(varchar(10),t.start_time_temp,120) = c.r and
CONVERT(varchar(10),t.end_time_temp,120) = c.r
then case when convert(varchar(5),t.start_time_temp,114) between '08:30' and '12:00'
and not (convert(varchar(5),t.end_time_temp,114) between '08:30' and '12:00')
then DATEDIFF(second,t.start_time_temp,c.r +' 12:00:00')
else 0
end +
case when convert(varchar(5),t.end_time_temp,114) between '14:00' and '18:00'
and not (convert(varchar(5),t.start_time_temp,114) between '14:00' and '18:00')
then DATEDIFF(second,c.r+' 14:00:00',t.end_time_temp)
else 0
end +
case when (convert(varchar(5),t.start_time_temp,114) between '08:30' and '12:00'
and convert(varchar(5),t.end_time_temp,114) between '08:30' and '12:00')
or
(convert(varchar(5),t.end_time_temp,114) between '14:00' and '18:00'
and convert(varchar(5),t.start_time_temp,114) between '14:00' and '18:00')
then DATEDIFF(SECOND,t.start_time_temp,t.end_time_temp)
else 0
end

/*
注意下面的计算逻辑是，如果这天不是假日，同时与开始日期相同
那么就要计算时间间隔，如果时间是在上午的工作时间范围内，
那么用当前日期的12点，减去开始日期，就是时间间隔，但还必须要加上下午的工作时间，
也就是4个小时，转化为秒数，就是4*3600
*/
when h_date IS null and
convert(varchar(10),t.start_time_temp,120) = c.r
then case when convert(varchar(5),t.start_time_temp,114) between '08:30' and '12:00'
then DATEDIFF(second,t.start_time_temp,c.r +' 12:00:00') + 4 * 3600
else 0
end +
case when convert(varchar(5),t.start_time_temp,114) between '14:00' and '18:00'
then DATEDIFF(second,t.start_time_temp,c.r +' 18:00:00')
else 0
end

when h_date IS null and
CONVERT(varchar(10),t.end_time_temp,120) = c.r
then case when convert(varchar(5),t.end_time_temp,114) between '08:30' and '12:00'
then DATEDIFF(second,c.r +' 08:30:00',t.end_time_temp)
else 0
end +
case when convert(varchar(5),t.end_time_temp,114) between '14:00' and '18:00'
then DATEDIFF(second,c.r +' 14:00:00',t.end_time_temp) + 3.5 * 3600
else 0
end

when h_date is null and
convert(varchar(10),t.start_time_temp,120) < c.r and
CONVERT(varchar(10),t.end_time_temp,120) > c.r
then 7.5 * 3600

when h_date IS null
then 0
end as seconds
FROM
(
/*
这里之所以要转换，是由于有些时间比如 start_time为2013-09-18 08:21:32.037,
不在正常工作时间（上午8:30--12:00 下午14:00--18:00）内，
所以要先转化为正常工作时间，否则后面的case when的逻辑判断就太复杂了。
*/
SELECT start_time,
end_time,

case when CONVERT(varchar(5),start_time,114) < '08:30'
then cast(CONVERT(varchar(10),start_time,120) + ' 08:30:00' AS datetime)

when CONVERT(varchar(5),start_time,114) between '12:00' and '14:00'
then cast(CONVERT(varchar(10),start_time,120) + ' 12:00:00' AS datetime)

else start_time
end as start_time_temp,

case when CONVERT(varchar(5),end_time,114) between '12:00' and '14:00'
then cast(CONVERT(varchar(10),end_time,120) + ' 12:00:00' AS datetime)

when CONVERT(varchar(5),end_time,114) > '18:00'
then cast(CONVERT(varchar(10),end_time,120) + ' 18:00:00' AS datetime)

else end_time
end as end_time_temp
FROM tab
) t
inner join calendar c
on convert(varchar(10),t.start_time,120) <= c.r
and convert(varchar(10),t.end_time,120) >= c.r
left join holiday h
on c.r = h.h_date

--OPTION(MAXRECURSION 1000) --限制最大递归次数
)

--select * from tt

select start_time,
end_time,

--汇总秒数，同时转化为小时
cast(round(SUM(seconds) / 3600 ,1,1) as numeric(10,1)) as diff_time
from tt
group by start_time,
end_time

/*
start_time end_time diff_time
2013-09-09 10:59:59.837 2013-09-09 14:06:21.223 1.1
2013-09-13 16:28:29.833 2013-09-16 09:41:47.987 2.7
2013-09-18 17:02:40.443 2013-09-22 15:27:58.983 5.9
2013-09-18 08:21:32.037 2013-09-22 15:31:52.500 12.5
2013-09-11 17:26:02.383 2013-09-24 10:38:01.410 55.1
*/