In Django, how do I check if a user is in a certain group?


I created a custom group in Django’s admin site.

In my code, I want to check if a user is in this group. How do I do that?

Asked By: TIMEX



You can access the groups simply through the groups attribute on User.

from django.contrib.auth.models import User, Group

group = Group(name = "Editor")                    # save this new group for this example
user = User.objects.get(pk = 1) # assuming, there is one initial user 
user.groups.add(group)          # user is now in the "Editor" group

then user.groups.all() returns [<Group: Editor>].

Alternatively, and more directly, you can check if a a user is in a group by:

if django_user.groups.filter(name = groupname).exists():


Note that groupname can also be the actual Django Group object.

Answered By: miku

If you need the list of users that are in a group, you can do this instead:

from django.contrib.auth.models import Group
users_in_group = Group.objects.get(name="group name").user_set.all()

and then check

 if user in users_in_group:
     # do something

to check if the user is in the group.

Update 2023

Looking at this solution 10 years later, I’m pretty sure that I would NOT want to ever to fetch a whole list of users like this. It’s something that would be problematic at scale. You would only want to fetch a list of users in a very specific use case where there were assurances that the list of users was going to remain small, or if you were just using the Django shell.

Answered By: Mark Chackerian

Just in case if you wanna check user’s group belongs to a predefined group list:

def is_allowed(user):
    allowed_group = set(['admin', 'lead', 'manager'])
    usr = User.objects.get(username=user)
    groups = [ for x in usr.groups.all()]
    if allowed_group.intersection(set(groups)):
       return True
    return False
Answered By: James Sapam

Your User object is linked to the Group object through a ManyToMany relationship.

You can thereby apply the filter method to user.groups.

So, to check if a given User is in a certain group ("Member" for the example), just do this :

def is_member(user):
    return user.groups.filter(name='Member').exists()

If you want to check if a given user belongs to more than one given groups, use the __in operator like so :

def is_in_multiple_groups(user):
    return user.groups.filter(name__in=['group1', 'group2']).exists()

Note that those functions can be used with the @user_passes_test decorator to manage access to your views :

from django.contrib.auth.decorators import login_required, user_passes_test

@user_passes_test(is_member) # or @user_passes_test(is_in_multiple_groups)
def myview(request):
    # Do your processing

For class based views, you might use UserPassesTestMixin with test_func method:

from django.contrib.auth.mixins import LoginRequiredMixin, UserPassesTestMixin

class MyView(LoginRequiredMixin, UserPassesTestMixin, View):

    login_url = '/login/'
    redirect_field_name = 'redirect_to'

    def test_func(self):
        return is_member(self.request.user)

Hope this help

Answered By: Charlesthk

In one line:

'Groupname' in user.groups.values_list('name', flat=True)

This evaluates to either True or False.

Answered By: Philipp Zedler

You just need one line:

from django.contrib.auth.decorators import user_passes_test  

@user_passes_test(lambda u: u.groups.filter(name='companyGroup').exists())
def you_view():
    return HttpResponse("Since you're logged in, you can see this text!")

If you don’t need the user instance on site (as I did), you can do it with

User.objects.filter(pk=userId, groups__name='Editor').exists()

This will produce only one request to the database and return a boolean.

Answered By: David Kühner

If a user belongs to a certain group or not, can be checked in django templates using:

{% if group in request.user.groups.all %}
"some action"
{% endif %}

Answered By: CODEkid

I have done it the following way. Seems inefficient but I had no other way in my mind:

def list_track(request):

usergroup = request.user.groups.values_list('name', flat=True).first()
if usergroup in 'appAdmin':
    tracks = QuestionTrack.objects.order_by('pk')
    return render(request, 'cmit/appadmin/list_track.html', {'tracks': tracks})

    return HttpResponseRedirect('/cmit/loggedin')
Answered By: Mohammad

User.objects.filter(username='tom', groups__name='admin').exists()

That query will inform you user : “tom” whether belong to group “admin ” or not

Answered By: Trung Lê

I did it like this. For group named Editor.

def index(request):
    current_user_groups = request.user.groups.values_list("name", flat=True)
    context = {
        "is_editor": "Editor" in current_user_groups,
    return render(request, "index.html", context)


# index.html
{% if is_editor %}
  <h1>Editor tools</h1>
{% endif %}
Answered By: Harry Moreno

I have similar situation, I wanted to test if the user is in a certain group. So, I’ve created new file where I put all my small utilities that help me through entire application. There, I’ve have this definition:

def is_company_admin(user):
    return user.groups.filter(name='company_admin').exists()

so basically I am testing if the user is in the group company_admin and for clarity I’ve called this function is_company_admin.

When I want to check if the user is in the company_admin I just do this:

from .utils import *

if is_company_admin(request.user):
        data = Company.objects.all().filter(

Now, if you wish to test same in your template, you can add is_user_admin in your context, something like this:

return render(request, 'admin/users.html', {'data': data, 'is_company_admin': is_company_admin(request.user)})

Now you can evaluate you response in a template:


{% if is_company_admin %}
     ... do something ...
{% endif %}

Simple and clean solution, based on answers that can be found earlier in this thread, but done differently. Hope it will help someone.

Tested in Django 3.0.4.

Answered By: Branko Radojevic

Use this:

{% for group in request.user.groups.all %}
    {% if == 'GroupName' %}
    {% endif %}
{% endfor %}
Answered By: rodolfodalvi