这里有一个快速而肮脏的黑客，只需获得你需要的信息，而不需要修改任何设置。

$ 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())'

这是小猪在逃避@robert的回答。虽然是正确的，但我不想让django扩展成为依赖项，哪怕只是一秒钟。

你可以创建一个动态导入来收集项目中每个应用程序的所有URL模式，简单的方法如下:

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()

我最近使用类似的方法创建了一个模板标记来显示活动导航链接。

试试这个:

from django.urls import get_resolver
get_resolver().reverse_dict.keys()

或者如果你还在Django 1上。*:

from django.core.urlresolvers import get_resolver
get_resolver(None).reverse_dict.keys()

Django >= 2.0列表解决方案

领养自@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

有一个我使用的插件:https://github.com/django-extensions/django-extensions，它有一个show_urls命令可以帮助。

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')
]