If you're looking for exact or "more precise" dates, you're probably better off checking out dateutil.
>>> from dateutil.relativedelta import relativedelta
>>> import datetime
>>> TODAY = datetime.date.today()
datetime.date(2012, 3, 6)
Now add 3 months to
TODAY, observe that it matches the day exactly (Note that
relativedelta(month=3) have different behaviors. Make sure to use
months for these examples!).
>>> three_mon_rel = relativedelta(months=3)
>>> TODAY + three_mon_rel
datetime.date(2012, 6, 6)
And it stays consistent throughout the course of a year. Literally every three months, on the day (had to keep adding because for some reason multiplying a
relativedelta and adding it to a
datetime.date object throws a
>>> TODAY + three_mon_rel + three_mon_rel
datetime.date(2012, 9, 6)
>>> TODAY + three_mon_rel + three_mon_rel + three_mon_rel
datetime.date(2012, 12, 6)
>>> TODAY + three_mon_rel + three_mon_rel + three_mon_rel + three_mon_rel
datetime.date(2013, 3, 6)
Whereas the mVChr's suggested solution, while definitely "good enough", drifts slightly over time:
>>> three_mon_timedelta = datetime.timedelta(days=3 * 365/12)
>>> TODAY + three_mon_timedelta
datetime.date(2012, 6, 5)
And over the course of a year, the day of month keeps sliding:
>>> TODAY + three_mon_timedelta * 2
datetime.date(2012, 9, 4)
>>> TODAY + three_mon_timedelta * 3
datetime.date(2012, 12, 4)
>>> TODAY + three_mon_timedelta * 4
datetime.date(2013, 3, 5)
Tested and works on SQL 2005 and 2008. Not sure if this works in 2012 and later.
The solution uses DATENAME instead of DATEPART
select datename(dw,getdate()) --Thursday
select datepart(dw,getdate()) --2
This is work in sql 2014 also.
I don't have a link to hand. Search your documentation for 'date arithmetic' and/or the 'interval' datatype.
Here's an example, which is not dissimilar to your english statement of what you want: