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 1.8, Python 2.7+ 只需在Shell中运行这些命令。Python manage.py shell并执行以下代码。
from django.conf.urls import RegexURLPattern, RegexURLResolver
from django.core import urlresolvers
urls = urlresolvers.get_resolver(None)
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)
其他回答
只要输入一个你知道不存在的url,服务器就会返回一个错误消息,其中包含一个url模式列表。
例如,如果你在http://localhost:8000/something上运行一个网站
输入
http://localhost:8000/something/blahNonsense,您的服务器将返回url搜索列表并在浏览器中显示它
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
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
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')
]