Overwrite Django permission and modify auth_permission - django

Question

I want to create the permissions in views dynamically instead of the defaults at models
I create a main class called CreatorView
from django.views.generic import View
from django.forms.models import modelform_factory
class CreatorView(View):
model = None
fields = None
exclude = None
form = None
page_title = ''
def create_form(self):
default =dict()
if self.fields == None:
if self.exclude == None:
default['fields'] = self.fields = self.model._meta.fileds
else:
default['exclude'] = self.exclude
else:
if self.exclude:
raise Exception('error')
default['fields'] = self.fields
return modelform_factory(self.model,**default)
def get(self,request,*args,**kwargs):
return render('','base.html')
def post(self,request,*args,**kwargs):
... and so on
the main urls is:
urlpatterns = [
path('%s/%s' % (cls.model._meta.app_label, cls.__name__.lower()), cls.as_view(),
name='%s/%s' % (cls.model._meta.app_label, cls.__name__.lower())) for cls
in CreatorView.__subclasses__()]
if I make inheritance from CreatorView then my class should create a page
for example:
class Login(CreatorView):
model = Users
""" my overwrite methods and actions """
class Configurations(CreatorView):
model = Configure
""" my overwrite methods and actions """
class Teachers(CreatorView):
model = Teachers
""" my overwrite methods and actions """
class Students(CreatorView):
model = Students
""" my overwrite methods and actions """
and so on
this code will create to me four pages
I want to create table semi to django content_type model to be like:
id
app_label
page
1
myapp
login
2
myapp
configurations
3
myapp
teachers
4
myapp
students
Can I modify auth_permission table in Django to make content_type foreign key from my content_type?
If I can how to prevent insert default permissions and make my insertion for default permission?

Answer 1

You can create Permissions manually according to docs here
from myapp.models import BlogPost
from django.contrib.auth.models import Permission
from django.contrib.contenttypes.models import ContentType
content_type = ContentType.objects.get_for_model(BlogPost)
permission = Permission.objects.create(
codename='can_publish',
name='Can Publish Posts',
content_type=content_type,
)
OR
An enhanced permission library which enables a logic-based permission system to handle complex permissions in Django.
here