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

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')
]

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

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
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

这里有一个快速而肮脏的黑客,只需获得你需要的信息,而不需要修改任何设置。

$ pip install django-extensions
$ python manage.py shell -c 'from django.core.management import call_command; from django_extensions.management.commands.show_urls import Command; call_command(Command())'

这是小猪在逃避@robert的回答。虽然是正确的,但我不想让django扩展成为依赖项,哪怕只是一秒钟。