Django : How can I see a list of urlpatterns?

Question

How can I see the current urlpatterns that \"reverse\" is looking in?

I\'m calling reverse in a view with an argument that I think should work, but doesn\'t. Any wa

Answer 1

I have extended Seti's command to show namespace, all url parts, auto-adjust column widths, sorted by (namespace,name): https://gist.github.com/andreif/263a3fa6e7c425297ffee09c25f66b20

import 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()

Note: column order is kept in Python 3.6 and one would need to use OrderedDict in older versions.

Update: A new version with OrderedDict now lives in django-