How do I raise a Response Forbidden in django

Question

I\'d like to do the following:

raise HttpResponseForbidden()

But I get the error:

exceptions must be old-style classes or d         


        

Answer 1

You can optionally supply a custom template named "403.html" to control the rendering of 403 HTTP errors.

As correctly pointed out by @dave-halter, The 403 template can only be used if you raise PermissionDenied

Below is a sample view used to test custom templates "403.html", "404.html" and "500.html"; please make sure to set DEBUG=False in project's settings or the framework will show a traceback instead for 404 and 500.

from django.http import HttpResponse
from django.http import Http404
from django.core.exceptions import PermissionDenied


def index(request):

    html = """


  
    

        
home
        
Raise Error 403
        
Raise Error 404
        
Raise Error 500
    
  

"""

    action = request.GET.get('action', '')
    if action == 'raise403':
        raise PermissionDenied
    elif action == 'raise404':
        raise Http404
    elif action == 'raise500':
        raise Exception('Server error')

    return HttpResponse(html)