python 格式日期时间，带有“st”、“nd”、“rd”、“th”（英文序数后缀），如 PHP 的“S”

Question

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 output 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.

Answer 1

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.

Answer 2

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')

Answer 3

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

Answer 4

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]

Answer 5

Here is the fancier code :

def date_extention(number):
    if number%10 == 1:
        return '%dst' % number
    if number%10 == 2:
        return '%dnd' % number
    if number%10 == 3:
        return '%drd' % number
    if (number%10 >= 4) or (number%10== 0):
        return '%dth' % number

By the way this works with any number so i recommend you make a module out of it. Especially if you make a lot of user inputed programs...

Answer 6

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')