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

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

其他回答

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 1.11, Python 2.7.6

cd to_your_django_project Python manage.py shell

然后粘贴以下代码。

from django.conf.urls import RegexURLPattern, RegexURLResolver
from django.core import urlresolvers
urls = urlresolvers.get_resolver()

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)

样例输出:

^django-admin/^$
^django-admin/^login/$
^django-admin/^logout/$
^django-admin/^password_change/$
^django-admin/^password_change/done/$
^django-admin/^jsi18n/$
^django-admin/^r/(?P<content_type_id>\d+)/(?P<object_id>.+)/$
^django-admin/^wagtailimages/image/^$
^django-admin/^wagtailimages/image/^add/$
^django-admin/^wagtailimages/image/^(.+)/history/$
^django-admin/^wagtailimages/image/^(.+)/delete/$
^django-admin/^wagtailimages/image/^(.+)/change/$
^django-admin/^wagtailimages/image/^(.+)/$
...

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

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]>

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()