Django handler500 as a Class Based View

Question

Why does this not work

handler500 = TemplateView.as_view(template_name="500.html")

I get the following exception:

Traceback (most recent call last):
  File "/usr/lib/python2.6/wsgiref/handlers.py", line 94, in run    
    self.finish_response()
  File "/usr/lib/python2.6/wsgiref/handlers.py", line 134, in finish_response
    for data in self.result:
  File "/home/hatem/projects/leadsift_app/.virtualenv/lib/python2.6/site-packages/django/template/response.py", line 117, in __iter__
    raise ContentNotRenderedError('The response content must be 'ContentNotRenderedError: The response content must be rendered before it can be iterated over.

I found this set of notes that describe that you are shooting yourself in the foot to use class based views there, why is that?

EDIT: I have ended up using this ... but I am still hoping someone out there would tell me how to get the original oneliner or similar working

class Handler500(TemplateView):
    template_name = "500.html"  
    @classmethod
    def as_error_view(cls):
        v = cls.as_view()
        def view(request):
            r = v(request)
            r.render()
            return r
        return view
handler500 = Handler500.as_error_view()

Torsten Engelbrecht · Accepted Answer · 2015-03-27 12:44:21Z

2

I think its actually quite simple (in Django 1.7 with Python 3.4):

views.py

from django.http import HttpResponse
from django.views.generic.base import View

class Custom500View(View):
    def dispatch(self, request, *args, **kwargs):
        return HttpResponse('My custom django 500 page')

urls.py

from .views import Custom500View

handler500 = Custom500View.as_view()

edited Mar 27, 2015 at 12:44

answered Mar 27, 2015 at 10:20

Torsten Engelbrecht

13.5k4 gold badges50 silver badges49 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Pykler Over a year ago

Did you actually test this out? If you see the first snippet in my question does exactly what you are saying (in one line instead of 3). It does not work because the error view requires a simple handler ... unless it was fixed in recent versions of Django your answer will not work.

Torsten Engelbrecht Over a year ago

I am actually using this (with Django 1.7). How I am using is slightly different, since my 500 view yields JSON instead of HTML. That's why I posted the answer like this instead of using TemplateView.

Torsten Engelbrecht Over a year ago

As a side note, I am also using Python 3.4. I see that the original error is raised in Python 2.6 within the wsgiref library. Maybe this is actually a Python issue, rather than Django.

Pykler Over a year ago

It shouldn't be related to the version of python, just the version of django. Basically the error view passes less parameters in than what the normal view expects.

Torsten Engelbrecht Over a year ago

Which Django version are you using then? Seems like this might be some case of "works on my machine". :P

|

pydanny · Accepted Answer · 2012-12-03 05:59:42Z

0

I would rather just use stock 500 templates with static HTML in vanilla Django then do anything with code. This is one toggle I believe should not be touched.

answered Dec 3, 2012 at 5:59

pydanny

8,3946 gold badges36 silver badges42 bronze badges

1 Comment

nicbou Over a year ago

Unless you have minimal context that's required to render the views (e.g. loading templates based on the device, having an error report form etc). That does not really constitute an answer.

Collectives™ on Stack Overflow

Django handler500 as a Class Based View

2 Answers 2

views.py

urls.py

6 Comments

1 Comment

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

views.py

urls.py

6 Comments

1 Comment

Linked

Related