[Django]-How to force a user logout in Django?

94đź‘Ť

Update:

Since Django 1.7, users are automatically logged-out when their password changes. On each request, the current password hash is compared to the value saved in their session and if doesn’t match, the user is logged-out.

So, a simple password update has the effect of logging the user out. You can then disable the account for login, or advise them to use the password reset feature to set a new password and log in again.

Original:

I don’t think there is a sanctioned way to do this in Django yet.

The user id is stored in the session object, but it is encoded. Unfortunately, that means you’ll have to iterate through all sessions, decode and compare…

Two steps:

First delete the session objects for your target user. If they log in from multiple computers they will have multiple session objects.

from django.contrib.sessions.models import Session
from django.contrib.auth.models import User

# grab the user in question 
user = User.objects.get(username='johndoe')

[s.delete() for s in Session.objects.all() if s.get_decoded().get('_auth_user_id') == user.id]

Then, if you need to, lock them out….

user.is_active = False
user.save()
👤Harold

69đź‘Ť

Although Harold’s answer works in this specific case, I can see at least two important issues with it:

  1. This solution can only be used with a database session engine. In other situations (cache, file, cookie) the Session model would not be used.
  2. When the number of sessions and users in database grows, this becomes quite inefficient.

To solve those issues, I suggest you take another approach at the problem. The idea is to store somewhere the date when the user was logged in for a given session, and the last time you requested a user to be logged out.

Then whenever someone access your site, if the logged in date is lower than the log out date, you can force-logout the user. As dan said, there’s no practical difference between logging out a user immediately or on his next request to your site.

Now, let’s see a possible implementation of this solution, for django 1.3b1. In three steps:

1. store in the session the last login date

Fortunately, Django auth system exposes a signal called user_logged_in. You just have to register that signals, and save the current date in the session. At the bottom of your models.py :

from django.contrib.auth.signals import user_logged_in
from datetime import datetime

def update_session_last_login(sender, user=user, request=request, **kwargs):
    if request:
        request.session['LAST_LOGIN_DATE'] = datetime.now()
user_logged_in.connect(update_session_last_login)

2. request a force logout for a user

We just need to add a field and a method to the User model. There’s multiple ways to achieve that (user profiles, model inheritance, etc.) each with pros and cons.

For the sake of simplicity, I’m gonna use model inheritance here, if you go for this solution, don’t forget to write a custom authentication backend.

from django.contrib.auth.models import User
from django.db import models
from datetime import datetime

class MyUser(User):
    force_logout_date = models.DateTimeField(null=True, blank=True)

    def force_logout(self):
        self.force_logout_date = datetime.now()
        self.save()

Then, if you want to force logout for user johndoe, you just have to:

from myapp.models import MyUser
MyUser.objects.get(username='johndoe').force_logout()

3. implement the check on access

Best way here is to use a middleware as dan suggested. This middleware will access request.user, so you need to put it after 'django.contrib.auth.middleware.AuthenticationMiddleware' in your MIDDLEWARE_CLASSES setting.

from django.contrib.auth import logout

class ForceLogoutMiddleware(object):
    def process_request(self, request):
        if request.user.is_authenticated() and request.user.force_logout_date and \
           request.session['LAST_LOGIN_DATE'] < request.user.force_logout_date:
            logout(request)

That should do it.


Notes

  • Be aware of the performance implication of storing an extra field for your users. Using model inheritance will add an extra JOIN. Using user profiles will add an extra query. Modifying directly the User is the best way performance wise, but it is still a hairy topic.

  • If you deploy that solution on an existing site, you will probably have some trouble with existing sessions, which won’t have the 'LAST_LOGIN_DATE' key. You can adapt a bit the middleware code to deal with that case :

     from django.contrib.auth import logout
    
     class ForceLogoutMiddleware(object):
         def process_request(self, request):
             if request.user.is_authenticated() and request.user.force_logout_date and \
                ( 'LAST_LOGIN_DATE' not in request.session or \
                  request.session['LAST_LOGIN_DATE'] < request.user.force_logout_date ):
                 logout(request)
    
  • In django 1.2.x, there is no user_logged_in signal. Fall back to overriding the login function:

     from django.contrib.auth import login as dj_login
     from datetime import datetime
    
     def login(request, user):
         dj_login(request, user)
         request.session['LAST_LOGIN_DATE'] = datetime.now()
    
👤Clément

51đź‘Ť

I needed something similar in my app. In my case, if a user was set to inactive, I wanted to make sure if the user was already logged in that they will be logged out and not able to continue to use the site. After reading this post, I came to the following solution:

from django.contrib.auth import logout

class ActiveUserMiddleware(object):
    def process_request(self, request):
        if not request.user.is_authenticated:
            return
        if not request.user.is_active:
           logout(request)

Just add this middleware in your settings and off you go. In the case of changing passwords, you could introduce a new field in the userprofile model that forces a user to logout, check for the value of the field instead of is_active above, and also unset the field when a user logs in. The latter can be done with Django’s user_logged_in signal.

7đź‘Ť

Perhaps, a bit of middleware that references a list of users who have been forced to log out. Next time the user tries to do anything, log them out then, redirects them, etc.

Unless of course, they need to be logged out immediately. But then again, they wouldn’t notice until they next tried to make a request anyway, so the above solution may just work.

👤dan

6đź‘Ť

This is in response to Balon’s query:

Yes, with around 140k sessions to iterate through I can see why Harold’s answer may not be as fast as you may like!

The way I would recommend is to add a model whose only two properties are foreign keys to User and Session objects. Then add some middleware that keeps this model up-to-date with current user sessions. I have used this sort of setup before; in my case, I borrowed the sessionprofile module from this Single Sign-On system for phpBB (see the source code in the “django/sessionprofile” folder) and this (I think) would suit your needs.

What you would end up with is some management function somewhere in your code like this (assuming the same code names and layout as in the sessionprofile module linked above):

from sessionprofile.models import SessionProfile
from django.contrib.auth.models import User

# Find all SessionProfile objects corresponding to a given username
sessionProfiles = SessionProfile.objects.filter(user__username__exact='johndoe')

# Delete all corresponding sessions
[sp.session.delete() for sp in sessionProfiles]

(I think this will also delete the SessionProfile objects, as from what I recall, Django’s default behaviour when an object referenced by a ForeignKey is deleted is to cascade it and also delete the object containing the ForeignKey, but if not then it is trivial enough to delete the contents of sessionProfiles when you are done.)

👤pythonian4000

3đź‘Ť

You can also use direct django function to do that, it will update and logs out all other sessions for the user, except the current one.

from django.contrib.auth import update_session_auth_hash
update_session_auth_hash(self.request, user)

Docs for update_session_auth_hash here.

👤Pankaj Sharma

2đź‘Ť

As Tony Abou-Assaleh, I also needed to log out users who were set to inactive, so I started by implementing his solution. After some time I found out that the middleware is forcing a DB query on all requests (to check if the user was blocked), and thus hurts performance on pages that doesn’t require login.

I have a custom user object and Django >= 1.7, so what I ended up doing is overriding its get_session_auth_hash function to invalidate the session when the user is inactive. A possible implementation is:

def get_session_auth_hash(self):
    if not self.is_active:
        return "inactive"
    return super(MyCustomUser, self).get_session_auth_hash()

For this to work, django.contrib.auth.middleware.SessionAuthenticationMiddleware should be in settings.MIDDLEWARE_CLASSES

👤Tzach

1đź‘Ť

As others stated, you can iterate over all sessions in DB, decode all of them, and delete those belonging to that user. But it’s slow, particularly if your site has high traffic and there are lots of sessions.

If you need a faster solution, you can use a session backend that lets you query and get the sessions of a specific user. In these session backends, Session has a foreign key to User, so you don’t need to iterate over all session objects:

Using these backends, deleting all sessions of a user can be done in a single line of code:

user.session_set.all().delete()

Disclaimer: I am the author of django-qsessions.

0đź‘Ť

from django.contrib.sessions.models import Session

deleting user session

[s.delete() for s in Session.objects.all() if s.get_decoded().get('_auth_user_hash') == user.get_session_auth_hash()]

-5đź‘Ť

Even I faced this issue. Few spammers from India keep posting about those Baba and Molvi for love solutions.

What I did is at the time of posting just inserted this code:

if request.user.is_active==False:
            return HttpResponse('You are banned on the site for spaming.')
👤Pulkit Sharma

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