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)
will return
2007-09-22 15:07:00.000
2007-09-22 15:00:00.000
The above just truncates the seconds and minutes, producing the results asked for in the question. As @OMG Ponies pointed 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)
and you'll get:
2007-09-22 15:08:00.000
2007-09-22 15:00:00.000
Before the date data type was added in SQL Server 2008, I would use the above method to truncate the time portion from a datetime to get only the date. The idea is 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 gives you the original date with the time set to 00:00:00.000:
select dateadd(day, @days, 0)
or more succinctly:
select dateadd(day, datediff(day, 0, @dt), 0)
Using a different datepart (e.g. hour, mi) will work accordingly.