How to get the last day of the month?

calendar.monthrange provides this information:

calendar.monthrange(year, month)
    Returns weekday of first day of the month and number of days in month, for the specified year and month.

>>> import calendar>>> calendar.monthrange(2002, 1)(1, 31)>>> calendar.monthrange(2008, 2)  # leap years are handled correctly(4, 29)>>> calendar.monthrange(2100, 2)  # years divisible by 100 but not 400 aren't leap years(0, 28)

so:

calendar.monthrange(year, month)[1]

seems like the simplest way to go.

If you don't want to import the calendar module, a simple two-step function can also be:

import datetimedef last_day_of_month(any_day):    # this will never fail    # get close to the end of the month for any day, and add 4 days 'over'    next_month = any_day.replace(day=28) + datetime.timedelta(days=4)    # subtract the number of remaining 'overage' days to get last day of current month, or said programattically said, the previous day of the first of next month    return next_month - datetime.timedelta(days=next_month.day)

Outputs:

>>> for month in range(1, 13):...     print last_day_of_month(datetime.date(2012, month, 1))...2012-01-312012-02-292012-03-312012-04-302012-05-312012-06-302012-07-312012-08-312012-09-302012-10-312012-11-302012-12-31

EDIT: See @Blair Conrad's answer for a cleaner solution

>>> import datetime>>> datetime.date(2000, 2, 1) - datetime.timedelta(days=1)datetime.date(2000, 1, 31)