declare @dt datetime
set @dt = '09-22-2007 15:07:38.850'
select dateadd(mi, datediff(mi, 0, @dt), 0)
select dateadd(hour, datediff(hour, 0, @dt), 0)
The above code will return this:
This will just truncate the seconds and minutes, producing the results asked for in the question. As @OMG point out, if you want to round up/down, then you can add half a minute or half an hour respectively, then truncate:
select dateadd(mi, datediff(mi, 0, dateadd(s, 30, @dt)), 0)
select dateadd(hour, datediff(hour, 0, dateadd(mi, 30, @dt)), 0)
Before the date data type was added in SQL Server 2008, you can use the above method to truncate the time portion from a datetime to get only the date. To determine the number of days between the datetime in question and a fixed point in time (0, which implicitly casts to 1900-01-01 00:00:00.000):
declare @days int
set @days = datediff(day, 0, @dt)
And then add that number of days to the fixed point in time, which will give you the original date with the time set to 00:00:00.000:
select dateadd(day, @days, 0)
In brief, you can use:
select dateadd(day, datediff(day, 0, @dt), 0)
Try using a different datepart (e.g. hour, mi) this will work.