Formatting “yesterday's” date in python

Question

I need to find \"yesterday\'s\" date in this format MMDDYY in Python.

So for instance, today\'s date would be represented like this: 111009

I ca

Answer 1

Use datetime.timedelta()

>>> from datetime import date, timedelta
>>> yesterday = date.today() - timedelta(days=1)
>>> yesterday.strftime('%m%d%y')
'110909'

Answer 2

To expand on the answer given by Chris

if you want to store the date in a variable in a specific format, this is the shortest and most effective way as far as I know

>>> from datetime import date, timedelta                   
>>> yesterday = (date.today() - timedelta(days=1)).strftime('%m%d%y')
>>> yesterday
'020817'

If you want it as an integer (which can be useful)

>>> yesterday = int((date.today() - timedelta(days=1)).strftime('%m%d%y'))
>>> yesterday
20817

Answer 3

Could I just make this somewhat more international and format the date according to the international standard and not in the weird month-day-year, that is common in the US?

from datetime import datetime, timedelta

yesterday = datetime.now() - timedelta(days=1)
yesterday.strftime('%Y-%m-%d')

Answer 4

This should do what you want:

import datetime
yesterday = datetime.datetime.now() - datetime.timedelta(days = 1)
print yesterday.strftime("%m%d%y")

Answer 5

all answers are correct, but I want to mention that time delta accepts negative arguments.

>>> from datetime import date, timedelta
>>> yesterday = date.today() + timedelta(days=-1)
>>> print(yesterday.strftime('%m%d%y')) #for python2 remove parentheses 

Answer 6

from datetime import datetime, timedelta

yesterday = datetime.now() - timedelta(days=1)
yesterday.strftime('%m%d%y')