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

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')
]

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']>,
...
]

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)

django 2.0的极简解决方案

例如,如果你正在寻找installed_apps的第一个应用程序上的url,你可以像这样访问它:

from django.urls import get_resolver
from pprint import pprint

pprint(
    get_resolver().url_patterns[0].url_patterns
)

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]>

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')
]