Convert datetime object to a String of date only in Python

Question

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'.

Answer 1

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

Answer 2

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.

Answer 3

Another option:

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

Answer 4

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'

Answer 5

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'

Answer 6

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

Answer 7

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'

Answer 8

The sexiest version by far is with format strings.

from datetime import datetime

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

Answer 9

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

Answer 10

end_date = "2021-04-18 16:00:00"
end_date_string = end_date.strftime("%Y-%m-%d")
print(end_date_string)

Answer 11

You can convert datetime to string.

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

Answer 12

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

Answer 13

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'

Answer 14

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

Answer 15

I have used this method to insert dates to JSON object

my_json_string = json.dumps({'date_of_birth': '''{}'''.format(date_of_birth)})