How to get a list of all users with a specific per

问题:

I want to get a list of all Django auth user with a specific permission group, something like this:

user_dict = {
    'queryset': User.objects.filter(permisson='blogger')
}

I cannot find out how to do this. How are the permissions groups saved in the user model?

回答1:

If you want to get list of users by permission, look at this variant:

from django.contrib.auth.models import User, Permission
from django.db.models import Q

perm = Permission.objects.get(codename='blogger')  
users = User.objects.filter(Q(groups__permissions=perm) | Q(user_permissions=perm)).distinct()

回答2:

This would be the easiest

from django.contrib.auth import models

group = models.Group.objects.get(name='blogger')
users = group.user_set.all()

回答3:

I think for group permissions, permissions are stored against groups, and then users have groups linked to them. So you can just resolve the user - groups relation.

e.g.

518$ python manage.py shell

(InteractiveConsole)
>>> from django.contrib.auth.models import User, Group
>>> User.objects.filter(groups__name='monkeys')
[<User: cms>, <User: dewey>]

回答4:

Based on @Glader's answer, this function wraps it up in a single query, and has been modified to algo get the superusers (as by definition, they have all perms):

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

def users_with_perm(perm_name):
    return User.objects.filter(
        Q(is_superuser=True) |
        Q(user_permissions__codename=perm_name) |
        Q(groups__permissions__codename=perm_name)).distinct()

# Example:
queryset = users_with_perm('blogger')

回答5:

Do not forget that specifying permission codename is not enough because different apps may reuse the same codename. One needs to get permission object to query Users correctly:

def get_permission_object(permission_str):
    app_label, codename = permission_str.split('.')
    return Permission.objects.filter(content_type__app_label=app_label, codename=codename).first()

def get_users_with_permission(permission_str, include_su=True):
    permission_obj = get_permission_object(permission_str)
    q = Q(groups__permissions=permission_obj) | Q(user_permissions=permission_obj)
    if include_su:
        q |= Q(is_superuser=True)
    return User.objects.filter(q).distinct()

Code with imports: https://github.com/Dmitri-Sintsov/django-jinja-knockout/blob/master/django_jinja_knockout/models.py

回答6:

Groups are many-to-many with Users (you see, nothing unusual, just Django models...), so the answer by cms is right. Plus this works both ways: having a group, you can list all users in it by inspecting user_set attribute.

回答7:

Based on @Augusto's answer, I did the following with a model manager and using the authtools library. This is in querysets.py:

from django.db.models import Q
from authtools.models import UserManager as AuthUserManager

class UserManager(AuthUserManager):
    def get_users_with_perm(self, perm_name):
        return self.filter(
                Q(user_permissions__codename=perm_name) |
                Q(groups__permissions__codename=perm_name)).distinct()

And then in models.py:

from django.db import models
from authtools.models import AbstractEmailUser
from .querysets import UserManager


class User(AbstractEmailUser):
   objects = UserManager()

回答8:

Try this:

User.objects.filter(groups__permissions = Permission.objects.get(codename='blogger'))

回答9:

$ python manage.py shell <<'EOF'
> from django.contrib.auth.models import User
> User.objects.filter(groups__name='blogger')
> EOF
...
(InteractiveConsole)
>>> >>> [<User: foo>, <User: bar>, <User: baz>, '...(remaining elements truncated)...']

(simplified from cms' answer, which I can't edit)