我该如何查看“reverse”正在查找的当前urlpatterns?
我在视图中调用带有参数的反向函数,但是它却不能正常工作。有没有办法检查哪些内容存在以及为什么我的模式不起作用?
我该如何查看“reverse”正在查找的当前urlpatterns?
我在视图中调用带有参数的反向函数,但是它却不能正常工作。有没有办法检查哪些内容存在以及为什么我的模式不起作用?
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
import urls
get_resolved_urls(urls.urlpatterns)
Django 2.0的极简解决方案
比如,如果你正在寻找已安装应用程序的第一个应用程序中的网址,则可以像这样访问它:
from django.urls import get_resolver
from pprint import pprint
pprint(
get_resolver().url_patterns[0].url_patterns
)
django.core.urlresolvers
导入get_resolver
,它也适用于Django 1.*。感谢Marcio! - Fernando Costa Bertoldiimport sys
from django.core.management import BaseCommand
from django.conf.urls import RegexURLPattern, RegexURLResolver
from django.core import urlresolvers
def collect_urls(urls=None, namespace=None, prefix=None):
if urls is None:
urls = urlresolvers.get_resolver()
_collected = []
prefix = prefix or []
for x in urls.url_patterns:
if isinstance(x, RegexURLResolver):
_collected += collect_urls(x, namespace=x.namespace or namespace,
prefix=prefix + [x.regex.pattern])
elif isinstance(x, RegexURLPattern):
_collected.append({'namespace': namespace or '',
'name': x.name or '',
'pattern': prefix + [x.regex.pattern],
'lookup_str': x.lookup_str,
'default_args': dict(x.default_args)})
else:
raise NotImplementedError(repr(x))
return _collected
def show_urls():
all_urls = collect_urls()
all_urls.sort(key=lambda x: (x['namespace'], x['name']))
max_lengths = {}
for u in all_urls:
for k in ['pattern', 'default_args']:
u[k] = str(u[k])
for k, v in list(u.items())[:-1]:
# Skip app_list due to length (contains all app names)
if (u['namespace'], u['name'], k) == \
('admin', 'app_list', 'pattern'):
continue
max_lengths[k] = max(len(v), max_lengths.get(k, 0))
for u in all_urls:
sys.stdout.write(' | '.join(
('{:%d}' % max_lengths.get(k, len(v))).format(v)
for k, v in u.items()) + '\n')
class Command(BaseCommand):
def handle(self, *args, **kwargs):
show_urls()
注意:在Python 3.6中,列顺序被保留,而在旧版本中需要使用
OrderedDict
。
更新:现在有一个包含OrderedDict的新版本存在于django-s软件包中:https://github.com/5monkeys/django-bananas/blob/master/bananas/management/commands/show_urls.py
Django 1.8,Python 2.7+ 只需在Shell中运行以下命令: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)
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')
]
subprocess.run()
允许您从Python运行命令行可执行文件。您可以传递show_urls
参数到manage.py,捕获输出,必须使用shell选项(默认情况下-据我所知),并将其保存到res
变量中。然后,url_list
是一个列表...它使用“生成器”结构来生成每个列表项。假设res输出带有\t
字符的编码URL,我们取一个列表项并将其拆分,然后取余数(第一个元素),即[item for line in ...]
。...
需要迭代所有URL并读取它们。 - Shmack源自 @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
这是对 @Cesar Canassa 的生成器魔法的又一次适应。您可以将其添加到 yourapp/management/commands/dumpurls.py
目录中,以便作为 management.py
的子命令使用。
注意:我添加了一行代码,以确保它仅过滤yourapp
的URL。如果需要其他URL,请相应地更新或删除该行。
management.py
的子命令部署路径: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))
(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
def get_url_patterns():
import importlib
from django.apps import apps
list_of_all_url_patterns = list()
for name, app in apps.app_configs.items():
# you have a directory structure where you should be able to build the correct path
# my example shows that apps.[app_name].urls is where to look
mod_to_import = f'apps.{name}.urls'
try:
urls = getattr(importlib.import_module(mod_to_import), "urlpatterns")
list_of_all_url_patterns.extend(urls)
except ImportError as ex:
# is an app without urls
pass
return list_of_all_url_patterns
list_of_all_url_patterns = get_url_patterns()
最近我使用类似的方法创建了一个模板标签,用于显示活动导航链接。
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
your_main_app
是指您的 settings.py 文件所在的应用程序名称。router.get_urls()
(在Django应用程序的urls.py
文件中)中的urlpatterns来打印特定路由器的所有URL模式。urls.py
文件,并将打印语句添加到文件底部,以便整个文件看起来像这样: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())
然后将模式打印出来,如下所示:
[<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']>,
...
]