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/$' [name='Auth-login']>,
...
]
其他回答
import subprocces
res = subprocess.run(
'python manage.py show_urls',
capture_output=True,
shell=True,
)
url_list = [
line.split('\t')[0]
for line in res.stdout.decode().split('\n')
]
Django 1.11, Python 2.7.6
cd to_your_django_project Python manage.py shell
然后粘贴以下代码。
from django.conf.urls import RegexURLPattern, RegexURLResolver
from django.core import urlresolvers
urls = urlresolvers.get_resolver()
def if_none(value):
if value:
return value
return ''
def print_urls(urls, parent_pattern=None):
for url in urls.url_patterns:
if isinstance(url, RegexURLResolver):
print_urls(url, if_none(parent_pattern) + url.regex.pattern)
elif isinstance(url, RegexURLPattern):
print(if_none(parent_pattern) + url.regex.pattern)
print_urls(urls)
样例输出:
^django-admin/^$
^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/^(.+)/$
...
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]>
def get_resolved_urls(url_patterns):
url_patterns_resolved = []
for entry in url_patterns:
if hasattr(entry, 'url_patterns'):
url_patterns_resolved += get_resolved_urls(
entry.url_patterns)
else:
url_patterns_resolved.append(entry)
return url_patterns_resolved
在python中manage.py shell
import urls
get_resolved_urls(urls.urlpatterns)
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。