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
Django >= 2.0解决方案
我测试了这篇文章中的其他答案,它们要么不能与Django 2一起工作。X,不完整或太复杂。因此,以下是我的看法:
from django.conf import settings
from django.urls import URLPattern, URLResolver
urlconf = __import__(settings.ROOT_URLCONF, {}, {}, [''])
def list_urls(lis, acc=None):
if acc is None:
acc = []
if not lis:
return
l = lis[0]
if isinstance(l, URLPattern):
yield acc + [str(l.pattern)]
elif isinstance(l, URLResolver):
yield from list_urls(l.url_patterns, acc + [str(l.pattern)])
yield from list_urls(lis[1:], acc)
for p in list_urls(urlconf.urlpatterns):
print(''.join(p))
该代码打印所有url,不像其他解决方案,它将打印完整的路径,而不仅仅是最后一个节点。例如:
admin/
admin/login/
admin/logout/
admin/password_change/
admin/password_change/done/
admin/jsi18n/
admin/r/<int:content_type_id>/<path:object_id>/
admin/auth/group/
admin/auth/group/add/
admin/auth/group/autocomplete/
admin/auth/group/<path:object_id>/history/
admin/auth/group/<path:object_id>/delete/
admin/auth/group/<path:object_id>/change/
admin/auth/group/<path:object_id>/
admin/auth/user/<id>/password/
admin/auth/user/
... etc, etc
其他回答
又一次改编自凯撒·卡纳萨的发电机魔法。这可以添加到你的应用程序的yourapp/management/commands/dumpurls.py目录中,这样它就可以作为management.py中的子命令来访问。
注意:我添加了一行,以确保它只过滤你的应用程序。如果需要其他url,则相应地更新或删除它。
作为management.py子命令
部署路径:yourapp/management/commands/dumpurls.py
from django.core.management.base import BaseCommand, CommandError
from django.conf import settings
from django.urls import URLPattern, URLResolver
def list_urls(lis, acc=None):
if acc is None:
acc = []
if not lis:
return
l = lis[0]
if isinstance(l, URLPattern):
yield acc + [str(l.pattern),l.name]
elif isinstance(l, URLResolver):
yield from list_urls(l.url_patterns, acc + [str(l.pattern)])
yield from list_urls(lis[1:], acc)
class Command(BaseCommand):
help = 'List all URLs from the urlconf'
def handle(self, *args, **options):
urlconf = __import__(settings.ROOT_URLCONF, {}, {}, [''])
records, glen, nlen = [], 0, 0
for p in list_urls(urlconf.urlpatterns):
record = [''.join(p[:2]), p[2]]
# Update me, or add an argument
if record[0].startswith('yourapp'):
clen = len(record[0])
if clen > glen: glen = clen
clen = len(record[1])
if clen > nlen: nlen = clen
records.append(record)
self.stdout.write('{:-<{width}}'.format('',width=glen+nlen))
self.stdout.write('{:<{glen}}Name'.format('Path',glen=glen+4))
self.stdout.write('{:-<{width}}'.format('',width=glen+nlen))
for record in records:
self.stdout.write('{path:<{glen}}{name}'.format(path=record[0],
name=record[1],
glen=glen+4))
self.stdout.write('{:-<{width}}'.format('',width=glen+nlen))
样例输出
(env) django@dev:myproj~> ./manage.py dumpurls
-------------------------------------------------------------------------------------------------------
Path Name
-------------------------------------------------------------------------------------------------------
yourapp/^api-key/$ api-key-list
yourapp/^api-key\.(?P<format>[a-z0-9]+)/?$ api-key-list
yourapp/^attacks/$ attack-list
yourapp/^attacks\.(?P<format>[a-z0-9]+)/?$ attack-list
yourapp/^attack-histories/$ attackhistory-list
yourapp/^attack-histories\.(?P<format>[a-z0-9]+)/?$ attackhistory-list
yourapp/^files/$ file-list
yourapp/^files\.(?P<format>[a-z0-9]+)/?$ file-list
yourapp/^modules/$ module-list
yourapp/^modules\.(?P<format>[a-z0-9]+)/?$ module-list
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()
from django.urls.resolvers import RegexPattern,RoutePattern
from your_main_app import urls
def get_urls():
url_list = []
for url in urls.urlpatterns:
url_list.append(url.pattern._regex) if isinstance(url.pattern, RegexPattern) else url_list.append(url.pattern._route)
return url_list
这里your_main_app是放置settings.py文件的应用程序名称
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)
Django >= 2.0解决方案
我测试了这篇文章中的其他答案,它们要么不能与Django 2一起工作。X,不完整或太复杂。因此,以下是我的看法:
from django.conf import settings
from django.urls import URLPattern, URLResolver
urlconf = __import__(settings.ROOT_URLCONF, {}, {}, [''])
def list_urls(lis, acc=None):
if acc is None:
acc = []
if not lis:
return
l = lis[0]
if isinstance(l, URLPattern):
yield acc + [str(l.pattern)]
elif isinstance(l, URLResolver):
yield from list_urls(l.url_patterns, acc + [str(l.pattern)])
yield from list_urls(lis[1:], acc)
for p in list_urls(urlconf.urlpatterns):
print(''.join(p))
该代码打印所有url,不像其他解决方案,它将打印完整的路径,而不仅仅是最后一个节点。例如:
admin/
admin/login/
admin/logout/
admin/password_change/
admin/password_change/done/
admin/jsi18n/
admin/r/<int:content_type_id>/<path:object_id>/
admin/auth/group/
admin/auth/group/add/
admin/auth/group/autocomplete/
admin/auth/group/<path:object_id>/history/
admin/auth/group/<path:object_id>/delete/
admin/auth/group/<path:object_id>/change/
admin/auth/group/<path:object_id>/
admin/auth/user/<id>/password/
admin/auth/user/
... etc, etc