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
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]>
其他回答
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)
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()
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]>
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 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)