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]>

有一个activestate的配方

import urls

def show_urls(urllist, depth=0):
    for entry in urllist:
        print("  " * depth, entry.regex.pattern)
        if hasattr(entry, 'url_patterns'):
            show_urls(entry.url_patterns, depth + 1)

show_urls(urls.url_patterns)

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()

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]>

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

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)