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
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
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')
]
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()
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)
有一个我使用的插件:https://github.com/django-extensions/django-extensions,它有一个show_urls命令可以帮助。