问题
Suppose I have a datetime field whose value is 2000-01-01 08:30:00 and a duration field whose value is say 00:15 (meaning 15 minutes)
If I subtract these two, I should get 2000-01-01 08:15:00
Also if I want to subtract 1:15 (means 1 hour 15 minutes), the output should be 2000-01-01 07:15:00
I am trying SELECT DATEDIFF(minute, '00:15','2000-01-01 08:30:00');
But the output is 52595055. How can i get the desired result?
N.B.~ If I do SELECT dateadd(minute, -15,'2000-01-01 08:30:00');
, I will get the desired result but that involves parsing the minute field.
Edit:
As per the answers, every one is suggesting converting everything into minutes and then to subtract - so if it is 1:30, i need to subtract 90 minutes. That's fine. Any other way without converting to minutes?
回答1:
SELECT DATEADD(minute, -15, '2000-01-01 08:30:00');
The second value (-15 in this case) must be numeric (i.e. not a string like '00:15'). If you need to subtract hours and minutes I would recommend splitting the string on the : to get the hours and minutes and subtracting using something like
SELECT DATEADD(minute, -60 * @h - @m, '2000-01-01 08:30:00');
where @h is the hour part of your string and @m is the minute part of your string
EDIT:
Here is a better way:
SELECT CAST('2000-01-01 08:30:00' as datetime) - CAST('00:15' AS datetime)
回答2:
You want to use DATEADD, using a negative duration. e.g.
DATEADD(minute, -15, '2000-01-01 08:30:00')
回答3:
Have you tried
SELECT DATEADD(mi, -15,'2000-01-01 08:30:00')
DATEDIFF is the difference between 2 dates.
回答4:
I spent a while trying to do the same thing, trying to subtract the hours:minutes
from datetime
- here's how I did it:
convert( varchar, cast((RouteMileage / @average_speed) as integer))+ ':' + convert( varchar, cast((((RouteMileage / @average_speed) - cast((RouteMileage / @average_speed) as integer)) * 60) as integer)) As TravelTime,
dateadd( n, -60 * CAST( (RouteMileage / @average_speed) AS DECIMAL(7,2)), @entry_date) As DepartureTime
OUTPUT:
DeliveryDate TravelTime DepartureTime
2012-06-02 12:00:00.000 25:49 2012-06-01 10:11:00.000
回答5:
Use DATEPART to pull apart your interval, and DATEADD to subtract the parts:
select dateadd(
hh,
-1 * datepart(hh, cast('1:15' as datetime)),
dateadd(
mi,
-1 * datepart(mi, cast('1:15' as datetime)),
'2000-01-01 08:30:00'))
or, we can convert to minutes first (though OP would prefer not to):
declare @mins int
select @mins = datepart(mi, cast('1:15' as datetime)) + 60 * datepart(hh, cast('1:15' as datetime))
select dateadd(mi, -1 * @mins, '2000-01-01 08:30:00')
来源:https://stackoverflow.com/questions/2309028/subtract-minute-from-datetime-in-sql-server-2005