SQL calculate Datediff for half of the 8 working day - sql

Question

I need to compare 2 dates and return the number of days in between. Here is a table as example:
+----+--------+-------------------------+-------------------------+
| id | userid | datestarted | datefinished |
+----+--------+-------------------------+-------------------------+
| | | | |
| 1 | 23 | 2014-03-25 09:05:00.000 | 2014-03-25 12:15:00.000 |
| 2 | 43 | 2014-03-25 09:05:00.000 | 2014-03-25 12:15:00.000 |
| 3 | 23 | 2014-03-31 09:05:00.000 | 2014-03-31 12:15:00.000 |
| 4 | 12 | 2014-03-25 09:05:00.000 | 2014-03-26 12:15:00.000 |
+----+--------+-------------------------+-------------------------+
In the first 3 cases, we have the same day, only the hours don't match.
Datestarted = 2014-03-25 09:05:00.000
Datefinished = 2014-03-25 12:15:00.000
We only input hours and minutes.
Until now, wee needed only to show the difference as whole number, without decimal points, and did it like this:
DATEDIFF(carsharing.datestarted, carsharing.datefinished)
But now, we have to show the difference between the dates as 0,5 day, if it is less than 4,5 hours. If the difference is greater it should stay as 1 day.
In the more complecated last case from the table, we should also compare and show difference between two different days
Datestarted = 2014-03-25 09:05:00.000
Datefinished = 2014-03-26 12:15:00.000
Here the result should be 1,5 days

Answer 1

I believe this is what you're looking for - this will round the difference to 0.5 for anything under 4.5 hours in the day, and everything else over that will go to a full day:
Declare #StartDate DateTime = '2014-03-25 09:05:00.000',
#EndDate DateTime = '2014-03-26 12:15:00.000'
;With TotalHours As
(
Select DateDiff(Minute, #StartDate, #EndDate) / 60.0 As TotalHours
)
Select Case
When TotalHours % 24 = 0
Then Floor(TotalHours / 24)
When TotalHours % 24 < 4.5
Then Floor(TotalHours / 24) + 0.5
Else Floor(TotalHours / 24) + 1.0
End As Days
From TotalHours

Answer 2

You can try this query. It gets the difference in minutes and multiply it by 2 in order to get a 0.5 day range. It then devide it by 24 hours and by 60 minutes before calculating the Ceiling value. Once you have it, it can be devide by 2 again.
When the value is over 4.5*24*60 (4.5 days in minutes), it only has to be devided by 24 and 60.
Query:
Select id, userid, datestarted, datefinished
, Days = Case When DATEDIFF(minute, datestarted, datefinished) > 4.5*60*24
then DATEDIFF(minute, datestarted, datefinished) / 24 / 60
else CEILING(((2.0*DATEDIFF(minute, datestarted, datefinished)) / 24 / 60)) / 2
end
From #dates
Output:
id userid datestarted datefinished Days
1 23 2014-03-25 09:05:00.000 2014-03-25 12:15:00.000 0.500000
2 43 2014-03-25 09:05:00.000 2014-03-25 12:15:00.000 0.500000
3 23 2014-03-31 09:05:00.000 2014-03-31 12:15:00.000 0.500000
4 12 2014-03-25 09:05:00.000 2014-03-26 12:15:00.000 1.500000
5 12 2014-03-25 09:05:00.000 2014-03-29 12:15:00.000 4.500000
6 12 2014-03-25 09:05:00.000 2014-03-29 22:15:00.000 4.000000
Sample Data
declare #dates table(id int, userid int, datestarted datetime, datefinished datetime);
insert into #dates(id, userid, datestarted, datefinished) values
(1, 23, '2014-03-25 09:05:00.000', '2014-03-25 12:15:00.000')
, (2, 43, '2014-03-25 09:05:00.000', '2014-03-25 12:15:00.000')
, (3, 23, '2014-03-31 09:05:00.000', '2014-03-31 12:15:00.000')
, (4, 12, '2014-03-25 09:05:00.000', '2014-03-26 12:15:00.000')
, (5, 12, '2014-03-25 09:05:00.000', '2014-03-29 12:15:00.000')
, (6, 12, '2014-03-25 09:05:00.000', '2014-03-29 22:15:00.000')

Answer 3

DECLARE
#StartDate datetime = '2014-03-25 09:05:00.000'
,#EndDate datetime = '2014-03-26 09:05:00.000'
;WITH d AS (SELECT DATEDIFF(d,#StartDate,#EndDate) Dys)
,h AS (SELECT DATEDIFF(hh,#StartDate,#EndDate) Hrs)
SELECT d.Dys + CASE WHEN (h.Hrs - d.Dys*24) = 0 THEN 0 ELSE CASE WHEN (h.Hrs - d.Dys*24) < 4.5 THEN 0.5 ELSE 1 END END
FROM d,h