Getting Django admin url for an object

Question

Before Django 1.0 there was an easy way to get the admin url of an object, and I had written a small filter that I\'d use like this:

Answer 1

from django.core.urlresolvers import reverse
def url_to_edit_object(obj):
  url = reverse('admin:%s_%s_change' % (obj._meta.app_label,  obj._meta.model_name),  args=[obj.id] )
  return u'<a href="%s">Edit %s</a>' % (url,  obj.__unicode__())

This is similar to hansen_j's solution except that it uses url namespaces, admin: being the admin's default application namespace.

Answer 2

If you are using 1.0, try making a custom templatetag that looks like this:

def adminpageurl(object, link=None):
    if link is None:
        link = object
    return "<a href=\"/admin/%s/%s/%d\">%s</a>" % (
        instance._meta.app_label,
        instance._meta.module_name,
        instance.id,
        link,
    )

then just use {% adminpageurl my_object %} in your template (don't forget to load the templatetag first)

Answer 3

There's another way for the later versions, for example in 1.10:

{% load admin_urls %}
<a href="{% url opts|admin_urlname:'add' %}">Add user</a>
<a href="{% url opts|admin_urlname:'delete' user.pk %}">Delete this user</a>

Where opts is something like mymodelinstance._meta or MyModelClass._meta

One gotcha is you can't access underscore attributes directly in Django templates (like {{ myinstance._meta }}) so you have to pass the opts object in from the view as template context.