[django] Getting Django admin url for an object

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: <a href="{{ object|admin_url }}" .... > ... </a>

Basically I was using the url reverse function with the view name being 'django.contrib.admin.views.main.change_stage'

reverse( 'django.contrib.admin.views.main.change_stage', args=[app_label, model_name, object_id] )

to get the url.

As you might have guessed, I'm trying to update to the latest version of Django, and this is one of the obstacles I came across, that method for getting the admin url doesn't work anymore.

How can I do this in django 1.0? (or 1.1 for that matter, as I'm trying to update to the latest version in the svn).

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.

You can use the URL resolver directly in a template, there's no need to write your own filter. E.g.

{% url 'admin:index' %}

{% url 'admin:polls_choice_add' %}

{% url 'admin:polls_choice_change' choice.id %}

{% url 'admin:polls_choice_changelist' %}

Ref: Documentation

I solved this by changing the expression to:

reverse( 'django-admin', args=["%s/%s/%s/" % (app_label, model_name, object_id)] )

This requires/assumes that the root url conf has a name for the "admin" url handler, mainly that name is "django-admin",

i.e. in the root url conf:

url(r'^admin/(.*)', admin.site.root, name='django-admin'),

It seems to be working, but I'm not sure of its cleanness.

Essentially the same as Mike Ramirez's answer, but simpler and closer in stylistics to django standard get_absolute_url method:

from django.urls import reverse

def get_admin_url(self):
    return reverse('admin:%s_%s_change' % (self._meta.app_label, self._meta.model_name),
                   args=[self.id])

Here's another option, using models:

Create a base model (or just add the admin_link method to a particular model)

class CommonModel(models.Model):
    def admin_link(self):
        if self.pk:
            return mark_safe(u'<a target="_blank" href="../../../%s/%s/%s/">%s</a>' % (self._meta.app_label,
                    self._meta.object_name.lower(), self.pk, self))
        else:
            return mark_safe(u'')
    class Meta:
        abstract = True

Inherit from that base model

   class User(CommonModel):
        username = models.CharField(max_length=765)
        password = models.CharField(max_length=192)

Use it in a template

{{ user.admin_link }}

Or view

user.admin_link()

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)

For pre 1.1 django it is simple (for default admin site instance):

reverse('admin_%s_%s_change' % (app_label, model_name), args=(object_id,))

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.