Getting Django admin url for an object

djangodjango-admindjango-urls

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).

Best Answer

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

Related Solutions

Python – Django Admin – change header ‘Django administration’ text

As of Django 1.7 you don't need to override templates. You can now implement site_header, site_title, and index_title attributes on a custom AdminSite in order to easily change the admin site’s page title and header text. Create an AdminSite subclass and hook your instance into your URLconf:

admin.py:

from django.contrib.admin import AdminSite
from django.utils.translation import ugettext_lazy

class MyAdminSite(AdminSite):
    # Text to put at the end of each page's <title>.
    site_title = ugettext_lazy('My site admin')

    # Text to put in each page's <h1> (and above login form).
    site_header = ugettext_lazy('My administration')

    # Text to put at the top of the admin index page.
    index_title = ugettext_lazy('Site administration')

admin_site = MyAdminSite()

urls.py:

from django.conf.urls import patterns, include
from myproject.admin import admin_site

urlpatterns = patterns('',
    (r'^myadmin/', include(admin_site.urls)),
)

Update: As pointed out by oxfn you can simply set the site_header in your urls.py or admin.py directly without subclassing AdminSite:

admin.site.site_header = 'My administration'

Override Django Admin URLs for Specific Model

Looking at the Django source, the admin URLs are built in two places, in the ModelAdmin instances, and in the AdminSite instances.

The part you want to change is built in the AdminSite instance (django.contrib.admin.sites.AdminSite), you can subclass that and override the get_urls method. If you look at the second half of the method you'll see this:

    # Add in each model's views.
    for model, model_admin in self._registry.iteritems():
        urlpatterns += patterns('',
            url(r'^%s/%s/' % (model._meta.app_label, model._meta.module_name),
                include(model_admin.urls))
        )

There it is adding the model's ._meta.module_name which is just the model's name lowercased (django.db.models.options.Options.contribute_to_class).

An easy way out is to override the Site's get_urls method and add a dict or special case for the Proxy model so it uses a different url instead of model._meta.module_name, something along the lines:

class MyAdminSite(AdminSite):

module_name_dict = {
    EventAdminProxy: 'myfunkymodulename'
}

def get_urls(self):
    base_patterns = super(MyAdminSite, self).get_urls()
    my_patterns = patterns('',)

    for model, model_admin in self._registry.iteritems():
        if model in self.module_name_dict:
            module_name = self.module_name_dict[model]
            my_patterns += patterns('',
                url(r'^%s/%s/' % (model._meta.app_label, module_name),
                    include(model_admin.urls))
            )

    return my_patterns + base_patterns

Best Answer

Related Solutions

Python – Django Admin – change header ‘Django administration’ text

Override Django Admin URLs for Specific Model

Related Topic