Django >= 2.0 solution

I tested the other answers in this post and they were either not working with Django 2.X, incomplete or too complex. Therefore, here is my take on this:

from django.conf import settings
from django.urls import URLPattern, URLResolver

urlconf = __import__(settings.ROOT_URLCONF, {}, {}, [''])

def list_urls(lis, acc=None):
    if acc is None:
        acc = []
    if not lis:
        return
    l = lis[0]
    if isinstance(l, URLPattern):
        yield acc + [str(l.pattern)]
    elif isinstance(l, URLResolver):
        yield from list_urls(l.url_patterns, acc + [str(l.pattern)])
    yield from list_urls(lis[1:], acc)

for p in list_urls(urlconf.urlpatterns):
    print(''.join(p))

This code prints all URLs, unlike some other solutions it will print the full path and not only the last node. e.g.:

admin/
admin/login/
admin/logout/
admin/password_change/
admin/password_change/done/
admin/jsi18n/
admin/r/<int:content_type_id>/<path:object_id>/
admin/auth/group/
admin/auth/group/add/
admin/auth/group/autocomplete/
admin/auth/group/<path:object_id>/history/
admin/auth/group/<path:object_id>/delete/
admin/auth/group/<path:object_id>/change/
admin/auth/group/<path:object_id>/
admin/auth/user/<id>/password/
admin/auth/user/
... etc, etc

In Django 3.0, it's as easy as:

from django.urls import get_resolver
print(get_resolver().url_patterns)

Prints: [<URLPattern '' [name='home']>, <URLPattern '/testing' [name='another_url']>]

Here is a quick and dirty hack to just get the information you need without needing to modify any of your settings.

$ pip install django-extensions
$ python manage.py shell -c 'from django.core.management import call_command; from django_extensions.management.commands.show_urls import Command; call_command(Command())'

This is piggy backing off @robert 's answer. While correct, I didn't want to have django-extensions as a dependency even if it was for just a second.

Simply type in a url you know does not exist and the server will return an error message with a list of url patterns.

For example, if you're running a site at http://localhost:8000/something

Type in

http://localhost:8000/something/blahNonsense, and your server will return the url search list and display it in the browser

Minimalist solution for django 2.0

For instance, if you're looking for an url that's on the first app of installed_apps, you can access it like that:

from django.urls import get_resolver
from pprint import pprint

pprint(
    get_resolver().url_patterns[0].url_patterns
)

Django 1.8, Python 2.7+ Just run these commands in your Shell. Python manage.py shell and execute the following code.

from django.conf.urls import RegexURLPattern, RegexURLResolver
from django.core import urlresolvers
urls = urlresolvers.get_resolver(None)

def if_none(value):
    if value:
        return value
    return ''

def print_urls(urls, parent_pattern=None):
    for url in urls.url_patterns:
        if isinstance(url, RegexURLResolver):
            print_urls(url, if_none(parent_pattern) + url.regex.pattern)
        elif isinstance(url, RegexURLPattern):
            print(if_none(parent_pattern) + url.regex.pattern)

print_urls(urls)

Django >= 2.0 List Solution

adopted from @CesarCanassa

from django.conf import settings
from django.urls import URLPattern, URLResolver

URLCONF = __import__(settings.ROOT_URLCONF, {}, {}, [''])

def list_urls(patterns, path=None):
    """ recursive """
    if not path:
        path = []
    result = []
    for pattern in patterns:
        if isinstance(pattern, URLPattern):
            result.append(''.join(path) + str(pattern.pattern))
        elif isinstance(pattern, URLResolver):
            result += list_urls(pattern.url_patterns, path + [str(pattern.pattern)])
    return result

You can create a dynamic import to gather all URL Patterns from each application in your project with a simple method like so:

def get_url_patterns():
    import importlib
    from django.apps import apps

    list_of_all_url_patterns = list()
    for name, app in apps.app_configs.items():
        # you have a directory structure where you should be able to build the correct path
        # my example shows that apps.[app_name].urls is where to look
        mod_to_import = f'apps.{name}.urls'
        try:
            urls = getattr(importlib.import_module(mod_to_import), "urlpatterns")
            list_of_all_url_patterns.extend(urls)
        except ImportError as ex:
            # is an app without urls
            pass

    return list_of_all_url_patterns

list_of_all_url_patterns = get_url_patterns()

I recently used something like this to create a template tag to show active navigation links.

from django.urls.resolvers import RegexPattern,RoutePattern
from your_main_app import urls

def get_urls():
    url_list = []
    for url in urls.urlpatterns:
        url_list.append(url.pattern._regex) if isinstance(url.pattern, RegexPattern) else url_list.append(url.pattern._route)

    return url_list

Here your_main_app is the app name where your settings.py file is placed

Yet another adaption of @Cesar Canassa 's generator magic. This can be added to the yourapp/management/commands/dumpurls.py director of your app so that it'll be accessible as a subcommand in management.py.

note: I added a line to make sure it filters for only yourapp. Update or remove it accordingly if additional URLs are desired.

As a management.py Subcommand

Deploy Path: yourapp/management/commands/dumpurls.py

from django.core.management.base import BaseCommand, CommandError
from django.conf import settings
from django.urls import URLPattern, URLResolver

def list_urls(lis, acc=None):

    if acc is None:
        acc = []
    if not lis:
        return
    l = lis[0]
    if isinstance(l, URLPattern):
        yield acc + [str(l.pattern),l.name]
    elif isinstance(l, URLResolver):
        yield from list_urls(l.url_patterns, acc + [str(l.pattern)])
    yield from list_urls(lis[1:], acc)

class Command(BaseCommand):
    help = 'List all URLs from the urlconf'

    def handle(self, *args, **options):

        urlconf = __import__(settings.ROOT_URLCONF, {}, {}, [''])

        records, glen, nlen = [], 0, 0

        for p in list_urls(urlconf.urlpatterns):
            record = [''.join(p[:2]), p[2]]

            # Update me, or add an argument
            if record[0].startswith('yourapp'):

                clen = len(record[0])
                if clen > glen: glen = clen

                clen = len(record[1])
                if clen > nlen: nlen = clen
                
                records.append(record)


        self.stdout.write('{:-<{width}}'.format('',width=glen+nlen))
        self.stdout.write('{:<{glen}}Name'.format('Path',glen=glen+4))
        self.stdout.write('{:-<{width}}'.format('',width=glen+nlen))
        for record in records:
            self.stdout.write('{path:<{glen}}{name}'.format(path=record[0],
                name=record[1],
                glen=glen+4))
        self.stdout.write('{:-<{width}}'.format('',width=glen+nlen))

Sample Output

(env) django@dev:myproj~> ./manage.py dumpurls
-------------------------------------------------------------------------------------------------------
Path                                                                        Name
-------------------------------------------------------------------------------------------------------
yourapp/^api-key/$                                                          api-key-list
yourapp/^api-key\.(?P<format>[a-z0-9]+)/?$                                  api-key-list
yourapp/^attacks/$                                                          attack-list
yourapp/^attacks\.(?P<format>[a-z0-9]+)/?$                                  attack-list
yourapp/^attack-histories/$                                                 attackhistory-list
yourapp/^attack-histories\.(?P<format>[a-z0-9]+)/?$                         attackhistory-list
yourapp/^files/$                                                            file-list
yourapp/^files\.(?P<format>[a-z0-9]+)/?$                                    file-list
yourapp/^modules/$                                                          module-list
yourapp/^modules\.(?P<format>[a-z0-9]+)/?$                                  module-list

import subprocces

res = subprocess.run(
    'python manage.py show_urls',
    capture_output=True,
    shell=True,
)
url_list = [
    line.split('\t')[0]
    for line in res.stdout.decode().split('\n')
]

In case you are using DRF, you can print all the URL patterns for a particular router by printing the urlpatterns from router.get_urls() (within your Django app's urls.py file).

Open your apps urls.py and add the print statement to the bottom of the file, so the whole file might look like this:

import pprint

from django.urls import include, path
from rest_framework import routers

from . import views

router = routers.DefaultRouter()
router.register(r"users", views.UserViewSet, basename="User")
router.register(r"auth", views.AuthenticationView, basename="Auth")
router.register(r"dummy", views.DummyViewSet, basename="Dummy")
router.register("surveys", views.SurveyViewSet, basename="survey")

urlpatterns = [
    path("", include(router.urls)),
]

pprint.pprint(router.get_urls())

The patterns are then printed like this:

[<URLPattern '^users/$' [name='User-list']>,
 <URLPattern '^users\.(?P<format>[a-z0-9]+)/?$' [name='User-list']>,
 <URLPattern '^users/admins/$' [name='User-admins']>,
 <URLPattern '^users/admins\.(?P<format>[a-z0-9]+)/?$' [name='User-admins']>,
 <URLPattern '^users/current/$' [name='User-current']>,
 <URLPattern '^users/current\.(?P<format>[a-z0-9]+)/?$' [name='User-current']>,
 <URLPattern '^users/(?P<pk>[^/.]+)/$' [name='User-detail']>,
 <URLPattern '^users/(?P<pk>[^/.]+)\.(?P<format>[a-z0-9]+)/?$' [name='User-detail']>,
 <URLPattern '^auth/login/$' [name='Auth-login']>,
...
]