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

你可以创建一个动态导入来收集项目中每个应用程序的所有URL模式,简单的方法如下:

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()

我最近使用类似的方法创建了一个模板标记来显示活动导航链接。

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')
]

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

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

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()
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)