Python - Date Format

Question

I have the code below that returns the date in the following format: m/d/yyyy, but I would like it to return it as mm/dd/yy( So it would show a 0- before single digit dates and only show the last two digits of the year). How can I tweak it so?

import datetime
today = datetime.date.today()
'{0.month}/{0.day}/{0.year}'.format(today)

score 1 · Answer 1 · answered May 29 '18 at 00:25

1

You're going to want to use strftime

datetime.date.today().strftime("%m/%d/%y")

You can find a list of directives here.

answered May 29 '18 at 00:25

ollien

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Or equivalently, if using `format` as above, `{0:%m/%d/%y}.format(today)` – donkopotamus May 29 '18 at 00:28
Hey, that's cool. I didn't know you could do that with the new format syntax! – ollien May 29 '18 at 00:30
@ donkopotamus Thank you! – 0004 May 29 '18 at 00:48

score 0 · Answer 2 · answered May 29 '18 at 00:49

0

import datetime
today = datetime.date.today()
'{0:%m%d%y}'.format(today).format(today)

returns: '052818'

answered May 29 '18 at 00:49

0004

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score 0 · Answer 3 · answered May 29 '18 at 01:48

Here is a sample program that gets you the output that you are seeking:

import datetime

now = datetime.datetime.now()

now = now.strftime('%m/%d/%y')
print(now)

We can use the strftime method to format the date. We pass the lowercase m (%m) because this will get return the month with a 0 in front of it. The lower case %d will also pass a day, and the lowercase %y will pass a year, only showing the last two digits as you specified

Here is your output:

05/28/18

And if you work with a day in the single digits, you get the 0 in front of the day that you wanted:

05/03/18

Python - Date Format

3 Answers3