[Django]-How to require login for Django Generic Views?

Django >= 1.9 or using django-braces

Django 1.9 has introduced a LoginRequiredMixin that is used thus:

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 pretty much the same mixin from django-braces – the Django version was based on the django-braces version. django-braces 1.4.x still supports Django 1.4 so you can use it with pretty old versions.

Older Methods

I found this question while googling for how to decorate class based views, so to add the answer for that:

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

Decorating 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())),
)

Decorating the class

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 to above for more details.

The generic views have changed from functions to objects with version 1.3 of Django. As such, there is a slight change needed for Will McCutchen and Will Hardy answers 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'))),
)

Also the documentation describes how to do this as well.

If you don’t want to write your own thin wrapper around the generic views in question (as Aamir 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
)

For django 1.11, You can use LoginRequiredMixin for Class-based Views

in settings file you should add

LOGIN_URL="/login/"

in your views.py

from django.contrib.auth.mixins import LoginRequiredMixin

class RestaurantLocationCreateView(LoginRequiredMixin,CreateView):
    ....

Another way to achieve this is below, I like that it is quite similar to how it’s done with function-based views and does not require modifying urls.py or overriding dispatch:

@method_decorator(login_required, name='dispatch')
class YourGenericViewSubclass(TemplateView):
    #
    # View methods
    #

In Django =>3.0 it gets pretty easy:

from django.contrib.auth.decorators import login_required
from django.utils.decorators import method_decorator
from django.views.generic import TemplateView

@method_decorator(login_required(login_url='/login/'), name='dispatch')
class ProtectedView(TemplateView):
    template_name = 'secret.html'

for reference: https://docs.djangoproject.com/en/3.0/topics/class-based-views/intro/#decorating-the-class

I wanted a re-usable way to require auth on many views derived from generic views. I created a replacement dispatch function which I can add to my view class in the same way as it’s 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

Use the following:

from django.contrib.auth.decorators import login_required

@login_required
def your_view():
    # your code here

The following could solve this issue.

// in views.py:
class LoginAuthenAJAX(View):
    def dispatch(self, request, *args, **kwargs):
        if request.user.is_authenticated:
            jsonr = json.dumps({'authenticated': True})
        else:
            jsonr = json.dumps({'authenticated': False})
        return HttpResponse(jsonr, content_type='application/json')

// in urls.py
    path('login_auth', views.LoginAuthenAJAX.as_view(), name="user_verify"),

//in xxx.html
<script src = “{% static “xxx/script.js” %}” 
var login_auth_link = “{%  url ‘user_verify’ %}”
</script>

// in script.js
        $.get(login_auth_link, {
            'csrfmiddlewaretoken' : csrf_token,
            },
            function(ret){
                if (ret.authenticated == false) {
                    window.location.pathname="/accounts/login/"
                }
                $("#message").html(ret.result);
            }
        )

I had been struggling with finding the answer to this for a long time till I found this workaround.

In models.py do:
from django.db import models

class YourLoginModel:
      fullname = models.CharField(max_length=255, default='your_name', unique=True)
      email  = models.EmailField(max_length=255, unique=True)
      username = models.CharField(max_length=255, unique=True)
      password = models.CharField(max_length=255) #using werkzeug's 
                                                  #generate_password_hash on plaintext password before committing to database model

In forms.py do:

from django import forms
from .models import YourLoginModel

class LoginForm(forms.ModelForm):
      class Meta:
            model = YourLoginModel
            fields = ('username', 'password')

In views.py login logic:

def login(request):
    #login logic here
     # init empty form
    form = LoginForm()

    if request.method == 'POST':

        try:
            # peforms a Select query in db and gets user with log in username
            user_logging_in = User.objects.get(username=request.POST['username'])

            # assign user hash to var
            hash = user_logging_in.password

            # assign form str passs word to var
            password = request.POST['password']

        # if the user does not exist
        except ObjectDoesNotExist:
            html_response = 'User does not exists'
            return HttpResponse(html_response)

        # peform password and hash check
        if check_password_hash(hash, password):
 
            #using sessions cookies to know who we're interacting with
            request.session['username'] = request.POST['username']

            #set expiry date of the session
            request.session.set_expiry(0) # 0 means when the browser is closed

            return redirect('yourapp:home')
        else:
            return HttpResponse('password was incorrect')

    html = 'Login'
    return render(request, 'login.html', {'form': form})

In app view you want to perform login_required on do

from django.views.generic import TemplateView

class yourTemplateView(TemplateView):
      template_name = 'your_template.html'
      def dispatch(self, request, *args, **kwrags):
           if not request.session.has_key('username'):
              #return HttpResponse('not logged in')
              return redirect('yourapp:login.html')
           else:
              return render(request, 'your_view.html')