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。

Django 1.8, Python 2.7+ 只需在Shell中运行这些命令。Python manage.py shell并执行以下代码。

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 1.11, Python 2.7.6

cd to_your_django_project Python manage.py shell

然后粘贴以下代码。

from django.conf.urls import RegexURLPattern, RegexURLResolver
from django.core import urlresolvers
urls = urlresolvers.get_resolver()

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)

样例输出:

^django-admin/^$
^django-admin/^login/$
^django-admin/^logout/$
^django-admin/^password_change/$
^django-admin/^password_change/done/$
^django-admin/^jsi18n/$
^django-admin/^r/(?P<content_type_id>\d+)/(?P<object_id>.+)/$
^django-admin/^wagtailimages/image/^$
^django-admin/^wagtailimages/image/^add/$
^django-admin/^wagtailimages/image/^(.+)/history/$
^django-admin/^wagtailimages/image/^(.+)/delete/$
^django-admin/^wagtailimages/image/^(.+)/change/$
^django-admin/^wagtailimages/image/^(.+)/$
...

Django >= 2.0解决方案

我测试了这篇文章中的其他答案，它们要么不能与Django 2一起工作。X，不完整或太复杂。因此，以下是我的看法:

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))

该代码打印所有url，不像其他解决方案，它将打印完整的路径，而不仅仅是最后一个节点。例如:

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

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。

在Django 3.0中，这很简单:

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

指纹: < urltern " >， < urltern '/测试"[另一个名称]>