Getting Django admin url for an object

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

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.

I had a similar issue where I would try to call reverse('admin_index') and was constantly getting django.core.urlresolvers.NoReverseMatch errors.

Turns out I had the old format admin urls in my urls.py file.

I had this in my urlpatterns:

(r'^admin/(.*)', admin.site.root),

which gets the admin screens working but is the deprecated way of doing it. I needed to change it to this:

(r'^admin/', include(admin.site.urls) ),

Once I did that, all the goodness that was promised in the Reversing Admin URLs docs started working.

Using template tag admin_urlname:

There's another way for the later versions (>=1.10), recommend by the Django documentation, using the template tag admin_urlname:

{% 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.

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])

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,))

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.

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)

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()