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

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']>,
...
]

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]>

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()

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。

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