python format datetime with "st", "nd", "rd", "th" (english ordinal suffix) like PHP's "S"

I would like a python datetime object to output (and use the result in django) like this:

Thu the 2nd at 4:30

But I find no way in python to o开发者_如何学Pythonutput st, nd, rd, or th like I can with PHP datetime format with the S string (What they call "English Ordinal Suffix") (http://uk.php.net/manual/en/function.date.php).

Is there a built-in way to do this in django/python? strftime isn't good enough (http://docs.python.org/library/datetime.html#strftime-strptime-behavior).

Django has a filter which does what I want, but I want a function, not a filter, to do what I want. Either a django or python function will be fine.

The django.utils.dateformat has a function format that takes two arguments, the first one being the date (a datetime.date [[or datetime.datetime]] instance, where datetime is the module in Python's standard library), the second one being the format string, and returns the resulting formatted string. The uppercase-S format item (if part of the format string, of course) is the one that expands to the proper one of 'st', 'nd', 'rd' or 'th', depending on the day-of-month of the date in question.

dont know about built in but I used this...

def ord(n):
    return str(n)+("th" if 4<=n%100<=20 else {1:"st",2:"nd",3:"rd"}.get(n%10, "th"))

and:

def dtStylish(dt,f):
    return dt.strftime(f).replace("{th}", ord(dt.day))

dtStylish can be called as follows to get Thu the 2nd at 4:30. Use {th} where you want to put the day of the month ("%d" python format code)

dtStylish(datetime(2019, 5, 2, 16, 30), '%a the {th} at %I:%M')

You can do this simply by using the humanize library

from django.contrib.humanize.templatetags.humanize import ordinal

You can then just give ordinal any integer, ie

ordinal(2) will return 2nd

I've just written a small function to solve the same problem within my own code:

def foo(myDate):
    date_suffix = ["th", "st", "nd", "rd"]

    if myDate % 10 in [1, 2, 3] and myDate not in [11, 12, 13]:
        return date_suffix[myDate % 10]
    else:
        return date_suffix[0]

My own version. Use python's strftime's format codes, substituting {th} where you want to see the ordinal suffix.

def format_date_with_ordinal(d, format_string):
    ordinal = {'1':'st', '2':'nd', '3':'rd'}.get(str(d.day)[-1:], 'th')
    return d.strftime(format_string).replace('{th}', ordinal)

or

format_date_with_ordinal = lambda d,f: d.strftime(f).replace('{th}', {'1':'st', '2':'nd', '3':'rd'}.get(str(d.day)[-1:], 'th'))

Following solution I got for above problem:

datetime.strptime(mydate, '%dnd %B %Y')
datetime.strptime(mydate, '%dst %B %Y')
datetime.strptime(mydate, '%dth %B %Y')
datetime.strptime(mydate, '%drd %B %Y')