How can I use Django permissions without defining a content type or model?

Question

I'd like to use a permissions based system to restrict certain actions within my Django application. These actions need not be related to a particular model (e.g. access to sections in the application, searching...), so I can't use the stock permissions framework directly, because the Permission model requires a reference to an installed content type.

I could write my own permission model but then I'd have to rewrite all the goodies included with the Django permissions, such as:

• The possibility to assign permissions to users and groups.
• The permission_required decorator.
• User.has_perm and related user methods.
• The perms template variable.
• ...

I've checked some apps like django-authority and django-guardian, but they seem to provide permissions even more coupled to the model system, by allowing per-object permissions.

Is there a way to reuse this framework without having defined any model (besides User and Group) for the project?

Answer 1

For those of you, who are still searching:

You can create an auxiliary model with no database table. That model can bring to your project any permission you need. There is no need to deal with ContentType or create Permission objects explicitly.

from django.db import models
        
class RightsSupport(models.Model):
            
    class Meta:
        
        managed = False  # No database table creation or deletion  \
                         # operations will be performed for this model. 
                
        default_permissions = () # disable "add", "change", "delete"
                                 # and "view" default permissions

        permissions = ( 
            ('customer_rights', 'Global customer rights'),  
            ('vendor_rights', 'Global vendor rights'), 
            ('any_rights', 'Global any rights'), 
        )

Right after manage.py makemigrations and manage.py migrate you can use these permissions like any other.

# Decorator

@permission_required('app.customer_rights')
def my_search_view(request):
    …

# Inside a view

def my_search_view(request):
    request.user.has_perm('app.customer_rights')

# In a template
# The currently logged-in user’s permissions are stored in the template variable {{ perms }}

{% if perms.app.customer_rights %}
    <p>You can do any customer stuff</p>
{% endif %}

---

EDIT: in case you are using a DB router, make sure to return True from allow_migrate for this model.

Answer 2

Fix for Chewie's answer in Django 1.8, which as been requested in a few comments.

It says in the release notes:

The name field of django.contrib.contenttypes.models.ContentType has been removed by a migration and replaced by a property. That means it’s not possible to query or filter a ContentType by this field any longer.

So it's the 'name' in reference in ContentType that the uses not in GlobalPermissions.

When I fix it I get the following:

from django.db import models
from django.contrib.auth.models import Permission
from django.contrib.contenttypes.models import ContentType


class GlobalPermissionManager(models.Manager):
    def get_queryset(self):
        return super(GlobalPermissionManager, self).\
            get_queryset().filter(content_type__model='global_permission')


class GlobalPermission(Permission):
    """A global permission, not attached to a model"""

    objects = GlobalPermissionManager()

    class Meta:
        proxy = True
        verbose_name = "global_permission"

    def save(self, *args, **kwargs):
        ct, created = ContentType.objects.get_or_create(
            model=self._meta.verbose_name, app_label=self._meta.app_label,
        )
        self.content_type = ct
        super(GlobalPermission, self).save(*args)

The GlobalPermissionManager class is unchanged but included for completeness.

If using Django admin to create/edit these permissions, think of removing the content_type field to avoid confusion:

class GlobalPermissionAdmin(admin.ModelAdmin):
    list_display = ("name", "codename")
    exclude = ("content_type",)

admin.site.register(GlobalPermission, GlobalPermissionAdmin)

Answer 3

Following Gonzalo's advice, I used a proxy model and a custom manager to handle my "modelless" permissions with a dummy content type.

from django.db import models
from django.contrib.auth.models import Permission
from django.contrib.contenttypes.models import ContentType


class GlobalPermissionManager(models.Manager):
    def get_queryset(self):
        return super(GlobalPermissionManager, self).\
            get_query_set().filter(content_type__name='global_permission')


class GlobalPermission(Permission):
    """A global permission, not attached to a model"""

    objects = GlobalPermissionManager()

    class Meta:
        proxy = True

    def save(self, *args, **kwargs):
        ct, created = ContentType.objects.get_or_create(
            name="global_permission", app_label=self._meta.app_label
        )
        self.content_type = ct
        super(GlobalPermission, self).save(*args, **kwargs)

Answer 4

I think that User model substitution do the trick (see Django documentation: https://docs.djangoproject.com/en/4.1/topics/auth/customizing/#substituting-a-custom-user-model)

Next on that model we can add Meta class with permissions and at the end we need to add our User model in settings.py.

Here is my example of user.py file:

from django.contrib.auth.models import AbstractUser

class User(AbstractUser):

    class Meta:
        default_permissions = []  # disable "add", "change", "delete"
        # and "view" default permissions

        permissions = [
            ('customer_rights', 'Global customer rights'),
            ('vendor_rights', 'Global vendor rights'), 
            ('any_rights', 'Global any rights'),
        ]

and in settings.py:

AUTH_USER_MODEL = 'myapp.User'

Answer 5

All answers are bad for me except this:

content_type = ContentType.objects.get_for_model(Permission)

Permission.objects.create(
    content_type=content_type,
    name='...', codename='...',
)

which handles model-less permissions without adding new models, but by adding new values.

Answer 6

For my part, for any larger project, I find it useful to have a generic app that isn't really part of my project's data model per se - I typically call it "projectlibs". It's a simple django app where I put things like fixtures for imports, templatetags that can be reused for multiple apps, etc. Some of it is template stuff I find myself re-using often, so the added benefit of having that type of stuff in an app is that it's reusable for other projects.

So inside that projectlibs/models.py, you could:

You could create that "meta app", in essence, and assign the content_type to some dummy class:

class UserRightsSupport(models.Model):
    class Meta:
        default_permissions = ()  # disable defaults add, delete, view, change perms
        permissions = (
            ("perm_name", "Verbose description"),
        )

Answer 7

You can use the proxy model for this with a dummy content type.

from django.contrib.auth.models import Permission
from django.contrib.contenttypes.models import ContentType


class CustomPermission(Permission):

    class Meta:
        proxy = True

    def save(self, *args, **kwargs):
        ct, created = ContentType.objects.get_or_create(
            model=self._meta.verbose_name, app_label=self._meta.app_label,
        )
        self.content_type = ct
        super(CustomPermission, self).save(*args)

Now you can create the permission with just name and codename of the permission from the CustomPermission model.

 CustomPermission.objects.create(name='Can do something', codename='can_do_something')

And you can query and display only the custom permissions in your templates like this.

 CustomPermission.objects.filter(content_type__model='custom permission')

Answer 8

This is alternative solution. First ask yourself: Why not create a Dummy-Model which really exists in DB but never ever gets used, except for holding permissions? That's not nice, but I think it is valid and straight forward solution.

from django.db import models

class Permissions(models.Model):

    can_search_blue_flower = 'my_app.can_search_blue_flower'

    class Meta:
        permissions = [
            ('can_search_blue_flower', 'Allowed to search for the blue flower'),
        ]

Above solution has the benefit, that you can use the variable Permissions.can_search_blue_flower in your source code instead of using the literal string "my_app.can_search_blue_flower". This means less typos and more autocomplete in IDE.

Answer 9

Django's Permission model requires a ContentType instance.

I think one way around it is creating a dummy ContentType that isn't related to any model (the app_label and model fields can be set to any string value).

If you want it all clean and nice, you can create a Permission proxy model that handles all the ugly details of the dummy ContentType and creates "modelless" permission instances. You can also add a custom manager that filters out all Permission instances related to real models.