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

Question

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?

Byssus

Answer 1

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

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

group = Group(name = "Editor")
group.save()                    # 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.

Babur

The actual checking would be if user.groups.filter(name=group_name).count(): # do something – Annecorinne

or use .exists() instead of .count() – Pinkham

The question is about query the User model for the groups it belongs, not how to instantiate'em... -.- – Shaunna

Answer 2

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

@login_required
@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

Alleenallegation

I'm not sure about the inner workings of django's DB access, but this seems a lot more efficient then some of the other suggestions, like getting all the users in a group and doing a standard python user in groups (or vice versa). – Nitride

Don't you have to add .exists() to the end to return a boolean? Otherwise is_member() and is_in_multiple_groups() will return a QuerySet, which may not give the desired result. – Smaragdite

According to Django's documentation, it is indeed faster to use exists() since it does not evaluate the queryset : docs.djangoproject.com/en/dev/ref/models/querysets/#exists – Alleenallegation

You probably want to have the superuser pass the test (without querying the database): def is_member(user): return user.is_superuser or user.groups.filter(... – Cutin

is_in_multiple_groups can be more explicitly named is_in_some_groups since it does not require that user be a member of all the groups – Hastings

how do you use this one in class based view?? – Trella

What file do you put this def in? I tried models.py and admin.py but neither of those seems right. – Launder

It always returns False in 2022 / Django 4.0 – Skirl

@DavidRhoden You could make a page in your app called functions.py and add it there, then in views.py or whatever python file you gonna use it if it is in that app, you just import it from the top like this from .functions import is_member . If you import it from another app, then from {thatAppName}.functions import is_member where {thatAppName} is the name of your app, for example if you named it app1, then from app1.functions import is_member – Skirl

If this happens to someone else. It returned False because I had lowercase, the first letter is capital letter, and I had lowercase. I saw this when I went to admin and checked what group this user was in. – Skirl

How do you add this to a Class Based View? – Skirl

I edited my answer to add an example for class based views – Alleenallegation

Answer 3

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.

Blaseio

Answer 4

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.

Rollway

This doesn't scale well for sites with more than a small number of users, since it will load large subsets users table into memory each time it runs. – Alcaraz

user.groups.filter(name="group name").exists() should work fine. The solution you have written uses two queries and therefore not very optimal. – Accompany

as it says, "if you need the list of users that are in a group"... – Rollway

indeed, a good solution to reduce db queries in order to check a bulk of users relationship with a given group – Explain

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 break at scale -- you would only want to use in a very specific use case where there where assurances that the list of users was going to remain small. – Rollway

Answer 5

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 %}

Thirza

this not work for me, seems that require compare group with groups name – Wicklund

Answer 6

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!")

Libbi

Not very clean code though, and not very reusable, but +1 for getting it into one line. – Chancemedley

Answer 7

Use this:

{% for group in request.user.groups.all %}
    {% if group.name == 'GroupName' %}
    {% endif %}
{% endfor %}

Juliojulis

I have been looking for this. Thanks – Aspirator

Answer 8

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

utils.py

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:

views.py

from .utils import *

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

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

views.py

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:

users.html

{% 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.

Outgrow

In your data = Company.objects.all().filter(id=request.user.company.id), what does Company signify? Is that your Model? – Cent

Yes @hayden, in this case Company is my model. – Outgrow

Answer 9

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 = [ x.name for x in usr.groups.all()]
    if allowed_group.intersection(set(groups)):
       return True
    return False

Freud

FYI you can say { x.name for x in usr.groups.all() } – Glyptograph

Answer 10

In one line:

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

This evaluates to either True or False.

Glennieglennis

This is inefficient, since it'll fetch a lot more data, and then operate on it on django's side. It better to use .exists() to let the db do the work. – Chancemedley

Answer 11

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

@login_required
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})

else:
    return HttpResponseRedirect('/cmit/loggedin')

Foreglimpse

Answer 12

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

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

Lilienthal

groups__name with double underscore – Donielle

Answer 13

I did it like this. For group named Editor.

# views.py
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)

template

# index.html
{% if is_editor %}
  <h1>Editor tools</h1>
{% endif %}

Millibar