Convert datetime object to a String of date only in Python

I see a lot on converting a date string to an datetime object in Python, but I want to go the other way.
I've got

datetime.datetime(2012, 2, 23, 0, 0)

and I would like to convert it to string like '2/23/2012'.

You can use strftime to help you format your date.

E.g.,

import datetime
t = datetime.datetime(2012, 2, 23, 0, 0)
t.strftime('%m/%d/%Y')

will yield:

'02/23/2012'

More information about formatting see here

date and datetime objects (and time as well) support a mini-language to specify output, and there are two ways to access it:

• direct method call: dt.strftime('format here')
• format method (python 2.6+): '{:format here}'.format(dt)
• f-strings (python 3.6+): f'{dt:format here}'

So your example could look like:

• dt.strftime('The date is %b %d, %Y')
• 'The date is {:%b %d, %Y}'.format(dt)
• f'The date is {dt:%b %d, %Y}'

In all three cases the output is:

> The date is Feb 23, 2012

For completeness' sake: you can also directly access the attributes of the object, but then you only get the numbers:

'The date is %s/%s/%s' % (dt.month, dt.day, dt.year)
# The date is 02/23/2012

The time taken to learn the mini-language is worth it.

---

For reference, here are the codes used in the mini-language:

• %a Weekday as locale’s abbreviated name.

• %A Weekday as locale’s full name.

• %w Weekday as a decimal number, where 0 is Sunday and 6 is Saturday.

• %d Day of the month as a zero-padded decimal number.

• %b Month as locale’s abbreviated name.

• %B Month as locale’s full name.

• %m Month as a zero-padded decimal number. 01, ..., 12

• %y Year without century as a zero-padded decimal number. 00, ..., 99

• %Y Year with century as a decimal number. 1970, 1988, 2001, 2013

• %H Hour (24-hour clock) as a zero-padded decimal number. 00, ..., 23

• %I Hour (12-hour clock) as a zero-padded decimal number. 01, ..., 12

• %p Locale’s equivalent of either AM or PM.

• %M Minute as a zero-padded decimal number. 00, ..., 59

• %S Second as a zero-padded decimal number. 00, ..., 59

• %f Microsecond as a decimal number, zero-padded on the left. 000000, ..., 999999

• %z UTC offset in the form +HHMM or -HHMM (empty if naive), +0000, -0400, +1030

• %Z Time zone name (empty if naive), UTC, EST, CST

• %j Day of the year as a zero-padded decimal number. 001, ..., 366

• %U Week number of the year (Sunday is the first) as a zero padded decimal number.

• %W Week number of the year (Monday is first) as a decimal number.

• %c Locale’s appropriate date and time representation.

• %x Locale’s appropriate date representation.

• %X Locale’s appropriate time representation.

• %% A literal '%' character.

Another option:

import datetime
now=datetime.datetime.now()
now.isoformat()
# ouptut --> '2016-03-09T08:18:20.860968'

You could use simple string formatting methods:

>>> dt = datetime.datetime(2012, 2, 23, 0, 0)
>>> '{0.month}/{0.day}/{0.year}'.format(dt)
'2/23/2012'
>>> '%s/%s/%s' % (dt.month, dt.day, dt.year)
'2/23/2012'

If you are looking for a simple way of datetime to string conversion and can omit the format. You can convert datetime object to str and then use array slicing.

In [1]: from datetime import datetime

In [2]: now = datetime.now()

In [3]: str(now)
Out[3]: '2019-04-26 18:03:50.941332'

In [5]: str(now)[:10]
Out[5]: '2019-04-26'

In [6]: str(now)[:19]
Out[6]: '2019-04-26 18:03:50'

But note the following thing. If other solutions will rise an AttributeError when the variable is None in this case you will receive a 'None' string.

In [9]: str(None)[:19]
Out[9]: 'None'

type-specific formatting can be used as well:

t = datetime.datetime(2012, 2, 23, 0, 0)
"{:%m/%d/%Y}".format(t)

Output:

'02/23/2012'

If you want the time as well, just go with

datetime.datetime.now().__str__()

Prints 2019-07-11 19:36:31.118766 in console for me

You can easly convert the datetime to string in this way:

from datetime import datetime

date_time = datetime(2012, 2, 23, 0, 0)
date = date_time.strftime('%m/%d/%Y')
print("date: %s" % date)

These are some of the patterns that you can use to convert datetime to string:

For better understanding, you can take a look at this article on how to convert strings to datetime and datetime to string in Python or the official strftime documentation

The sexiest version by far is with format strings.

from datetime import datetime

print(f'{datetime.today():%Y-%m-%d}')

You can convert datetime to string.

published_at = "{}".format(self.published_at)

It is possible to convert a datetime object into a string by working directly with the components of the datetime object.

from datetime import date  

myDate = date.today()    
#print(myDate) would output 2017-05-23 because that is today
#reassign the myDate variable to myDate = myDate.month 
#then you could print(myDate.month) and you would get 5 as an integer
dateStr = str(myDate.month)+ "/" + str(myDate.day) + "/" + str(myDate.year)    
# myDate.month is equal to 5 as an integer, i use str() to change it to a 
# string I add(+)the "/" so now I have "5/" then myDate.day is 23 as
# an integer i change it to a string with str() and it is added to the "5/"   
# to get "5/23" and then I add another "/" now we have "5/23/" next is the 
# year which is 2017 as an integer, I use the function str() to change it to 
# a string and add it to the rest of the string.  Now we have "5/23/2017" as 
# a string. The final line prints the string.

print(dateStr)  

Output --> 5/23/2017

String concatenation, str.join, can be used to build the string.

d = datetime.now()
'/'.join(str(x) for x in (d.month, d.day, d.year))
'3/7/2016'

end_date = "2021-04-18 16:00:00"

end_date_string = end_date.strftime("%Y-%m-%d")

print(end_date_string)

An approach to how far from now

• support different languages by passing in param li, a list corresponding timestamp.

from datetime import datetime
from dateutil import parser

t1 = parser.parse("Tue May 26 15:14:45 2021")
t2 = parser.parse("Tue May 26 15:9:45 2021")
# 5min
t3 = parser.parse("Tue May 26 11:14:45 2021")
# 4h
t4 = parser.parse("Tue May 26 11:9:45 2021")
# 1day
t6 = parser.parse("Tue May 25 11:14:45 2021")
# 1day4h
t7 = parser.parse("Tue May 25 11:9:45 2021")
# 1day4h5min
t8 = parser.parse("Tue May 19 11:9:45 2021")
# 1w
t9 = parser.parse("Tue Apr 26 11:14:45 2021")
# 1m
t10 = parser.parse("Tue Oct 08 06:00:20 2019") 
# 1y7m, 19m
t11 = parser.parse("Tue Jan 08 00:00:00 2019") 
# 2y4m, 28m


# create: date of object creation
# now: time now
# li: a list of string indicate time (in any language)
# lst: suffix (in any language)
# long: display length
def howLongAgo(create, now, li, lst, long=2):
	dif = create - now
	print(dif.days)
	sec = dif.days * 24 * 60 * 60 + dif.seconds
	minute = sec // 60
	sec %= 60
	hour = minute // 60
	minute %= 60
	day = hour // 24
	hour %= 24
	week = day // 7
	day %= 7
	month = (week * 7) // 30
	week %= 30
	year = month // 12
	month %= 12
	s = []
	for ii, tt in enumerate([sec, minute, hour, day, week, month, year]):
		ss = li[ii]
		if tt != 0:
			if tt == 1:
				s.append(str(tt) + ss)
			else:
				s.append(str(tt) + ss + 's')

	return ' '.join(list(reversed(s))[:long]) + ' ' + lst



t = howLongAgo(t1, t11, [
	'second', 
	'minute',
	'hour', 
	'day',
	'week', 
	'month',
	'year',
], 'ago')
print(t)
# 2years 4months ago