192

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?

Mark Chackerian
  • 21,866
  • 6
  • 108
  • 99
TIMEX
  • 259,804
  • 351
  • 777
  • 1,080

13 Answers13

270

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

Charlesthk
  • 9,394
  • 5
  • 43
  • 45
  • 4
    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). – brianmearns Dec 12 '13 at 15:32
  • 1
    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. – Michael Bates Sep 17 '14 at 06:34
  • 4
    According to Django's documentation, it is indeed faster to use exists() since it does not evaluate the queryset : https://docs.djangoproject.com/en/dev/ref/models/querysets/#exists – Charlesthk Sep 25 '14 at 21:10
  • 7
    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(...` – Dave Apr 27 '16 at 19:27
  • `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 – PeterVermont Jan 31 '17 at 21:58
  • how do you use this one in class based view?? – ramya Jul 15 '21 at 22:30
  • What file do you put this `def` in? I tried models.py and admin.py but neither of those seems right. – David Rhoden Oct 20 '21 at 15:51
  • It always returns **False** in 2022 / Django 4.0 – AnonymousUser Jun 20 '22 at 07:35
  • 1
    @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` – AnonymousUser Jun 20 '22 at 07:40
  • 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. – AnonymousUser Jun 20 '22 at 07:50
  • How do you add this to a **Class Based View**? – AnonymousUser Oct 07 '22 at 06:41
  • I edited my answer to add an example for class based views – Charlesthk Oct 07 '22 at 06:53
143

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.

Philippe Fanaro
  • 6,148
  • 6
  • 38
  • 76
miku
  • 181,842
  • 47
  • 306
  • 310
20

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.

David Kühner
  • 793
  • 1
  • 7
  • 19
16

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.

Mark Chackerian
  • 21,866
  • 6
  • 108
  • 99
  • 7
    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. – bhuber Oct 08 '13 at 16:17
  • 1
    `user.groups.filter(name="group name").exists()` should work fine. The solution you have written uses two queries and therefore not very optimal. – Noopur Phalak May 22 '19 at 09:18
  • as it says, "if you need the list of users that are in a group"... – Mark Chackerian May 29 '20 at 12:42
  • indeed, a good solution to reduce db queries in order to check a bulk of users relationship with a given group – Trigremm Jan 13 '23 at 09:19
  • 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. – Mark Chackerian Jan 13 '23 at 15:23
11

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 %}

CODEkid
  • 240
  • 4
  • 8
10

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!")
8

Use this:

{% for group in request.user.groups.all %}
    {% if group.name == 'GroupName' %}
    {% endif %}
{% endfor %}
rodolfodalvi
  • 81
  • 1
  • 1
4

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.

Branko Radojevic
  • 660
  • 1
  • 5
  • 14
1

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
James Sapam
  • 16,036
  • 12
  • 50
  • 73
  • 2
    [FYI you can say { x.name for x in usr.groups.all() }](http://docs.python.org/2/tutorial/datastructures.html#sets) – Kos Mar 21 '14 at 14:06
0

In one line:

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

This evaluates to either True or False.

Philipp Zedler
  • 1,660
  • 1
  • 17
  • 36
  • 4
    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. – WhyNotHugo Aug 25 '15 at 11:41
0

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')
Mohammad
  • 657
  • 1
  • 7
  • 18
0

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

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

Harry Moreno
  • 10,231
  • 7
  • 64
  • 116
Trung Lê
  • 31
  • 1
0

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 %}
Harry Moreno
  • 10,231
  • 7
  • 64
  • 116