How to require login to Django Generic Views?

Question

I want to restrict access to URLs handled by Django Generic Views.

For my views, I know that login_required decorator does the job. Also, Create / Delete / Update Generic Views accepts the login_required argument, but I could not find a way to do this for other general views.

Answer 1

For Django <1.5, you can add a decorator by wrapping the function in your URLs, which allows you to wrap general views:

 from django.contrib.auth.decorators import login_required from django.views.generic.simple import direct_to_template urlpatterns = patterns('', (r'^foo/$', login_required(direct_to_template), {'template': 'foo_index.html'}), ) 

Function-based common views are deprecated in Django 1.4 and removed in Django 1.5. But the same principle applies, just wrap the view-based class view function with the login_required decorator:

 login_required(TemplateView.as_view(template_name='foo_index.html')) 

Answer 2

Django 1.9 or using django-braces

Django 1.9 introduced LoginRequiredMixin , which is used this way:

 from django.contrib.auth.mixins import LoginRequiredMixin class MyView(LoginRequiredMixin, View): login_url = '/login/' redirect_field_name = 'redirect_to' 

If you are using an older version of django, you can use almost the same mixin from django-braces - the version of Django was based on the version of django-braces. django-braces 1.4.x still supports Django 1.4 , so you can use it with fairly old versions.

Old methods

I found this question during a search in order to decorate class based views, so add an answer for this:

This is described in the documentation section decorating class-based views . There is a urls.py , or you can apply a decorator to the dispatch() method. Examples from the documentation:

Decoration in URL conf

 from django.contrib.auth.decorators import login_required, permission_required from django.views.generic import TemplateView from .views import VoteView urlpatterns = patterns('', (r'^about/', login_required(TemplateView.as_view(template_name="secret.html"))), (r'^vote/', permission_required('polls.can_vote')(VoteView.as_view())), ) 

Class decoration

 from django.contrib.auth.decorators import login_required from django.utils.decorators import method_decorator from django.views.generic import TemplateView class ProtectedView(TemplateView): template_name = 'secret.html' @method_decorator(login_required) def dispatch(self, *args, **kwargs): return super(ProtectedView, self).dispatch(*args, **kwargs) 

See the documentation linked above for more details.

Answer 3

General views have changed from functions to objects with version 1.3 of Django. Thus, Will McCutchen and Will Hardy require a small change to work with version 1.3:

 from django.contrib.auth.decorators import login_required from django.views.generic import TemplateView urlpatterns = patterns('', (r'^foo/$', login_required(TemplateView.as_view(template_name='foo_index.html'))), ) 

The documentation also describes how to do this.

Answer 4

If you don't want to write your own thin shell around the general representations in question (as Amir suggested), you can also do something like this in your urls.py file:

 from django.conf.urls.defaults import * # Directly import whatever generic views you're using and the login_required # decorator from django.views.generic.simple import direct_to_template from django.contrib.auth.decorators import login_required # In your urlpatterns, wrap the generic view with the decorator urlpatterns = patterns('', (r'', login_required(direct_to_template), {'template': 'index.html'}), # etc ) 

Answer 5

For django 1.11 you can use LoginRequiredMixin for class based views

in the settings file you should add

 LOGIN_URL="/login/" 

in your views.py

 from django.contrib.auth.mixins import LoginRequiredMixin class RestaurantLocationCreateView(LoginRequiredMixin,CreateView): .... 

Answer 6

I need a reusable way to require authorization in many views derived from general views. I created a dispatch dispatch function that I can add to my view class just like other declarations.

 class Index(generic.ListView): model = models.HomePage dispatch = auth.dispatch 

auth.dispatch is where we do the work:

 def dispatch(self, request, *args, **kw): """Mix-in for generic views""" if userSession(request): return super(self.__class__, self).dispatch(request, *args, **kw) # auth failed, return login screen response = user(request) response.set_cookie('afterauth', value=request.path_info) return response 

Answer 7

Use the following:

 from django.contrib.auth.decorators import login_required @login_required def your_view(): # your code here