CODEWITHSUNDEEP
python-3.x
0004
0004

Reputation: 1260

Python - Subtract 3 months from Today's Date while keeping specificied date format

I currently have this code to print today's date in a format I need for my program, but can't figure out how to subtract 3 months (considering some months vary in days) and return said new date in the format in place below: 

import datetime
now = datetime.date.today() # create the date for today
today = '{0:%m%d%y}'.format(now).format(now)

Upvotes: 0

Views: 1162

Answers (1)

dawg
dawg

Reputation: 103844

Given:

>>> import datetime
>>> now = datetime.date.today() # create the date for today
>>> today = '{0:%m%d%y}'.format(now)
>>> now
datetime.date(2018, 7, 13)
>>> today
'071318'

You can use calendar:

import calendar
def monthdelta(date, delta):
     m, y = (date.month+delta) % 12, date.year + ((date.month)+delta-1) // 12
     if not m: m = 12
     d = min(date.day, calendar.monthrange(y, m)[1])
     return date.replace(day=d,month=m, year=y)

>>> '{0:%m%d%y}'.format(monthdelta(now,-3))
'041318'

This is Python 3 only because of //

Upvotes: 2

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