我怎么能看到当前的urlpatterns“反向”正在寻找?
我在一个视图中调用了逆向,我认为这个论证应该成立,但实际上并不成立。有什么办法能让我知道为什么我的图案没有?
我怎么能看到当前的urlpatterns“反向”正在寻找?
我在一个视图中调用了逆向,我认为这个论证应该成立,但实际上并不成立。有什么办法能让我知道为什么我的图案没有?
当前回答
如果你正在使用DRF,你可以通过从router.get_urls()(在你的Django应用的urls.py文件中)打印特定路由器的urlpatterns来打印所有的URL模式。
打开你的apps urls.py,并在文件底部添加打印语句,这样整个文件看起来就像这样:
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())
然后,这些图案会像这样打印出来:
[<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']>,
...
]
其他回答
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)
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。
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
如果你正在使用DRF,你可以通过从router.get_urls()(在你的Django应用的urls.py文件中)打印特定路由器的urlpatterns来打印所有的URL模式。
打开你的apps urls.py,并在文件底部添加打印语句,这样整个文件看起来就像这样:
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())
然后,这些图案会像这样打印出来:
[<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']>,
...
]
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。