[Django]-For a django model, how can I get the django admin URL to add another, or list objects, etc.?

by

64👍

✅

http://docs.djangoproject.com/en/dev/ref/contrib/admin/#reversing-admin-urls

obj = coconut_transportation.swallow.objects.all()[34]

# list url
url = reverse("admin:coconut_transportation_swallow_changelist")

# change url
url = reverse("admin:coconut_transportation_swallow_change", args=[obj.id])

# add url
url = reverse("admin:coconut_transportation_swallow_add")
👤Yuji 'Tomita' Tomita

15👍

You can retrieve this from the actual object instance, this worked for me:

from django.core import urlresolvers
from django.contrib.contenttypes.models import ContentType

content_type = ContentType.objects.get_for_model(object.__class__)
object_admin_url = django.core.urlresolvers.reverse("admin:%s_%s_change" % (content_type.app_label, content_type.model), args=(object.pk,))

See this: http://djangosnippets.org/snippets/1916/

👤Lior Boord

11👍

You can actually retrieve the info without making a query to ContentTypes

Just add this to your model class:

def get_admin_url(self):
    return urlresolvers.reverse("admin:%s_%s_change" %
        (self._meta.app_label, self._meta.model_name), args=(self.pk,))
👤bformet

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