Under your main views.py add your own custom implementation of the following two views, and just set up the templates 404.html and 500.html with what you want to display.
With this solution, no custom code needs to be added to urls.py
Here's the code:
from django.shortcuts import render_to_response
from django.template import RequestContext
def handler404(request, *args, **argv):
response = render_to_response('404.html', {},
context_instance=RequestContext(request))
response.status_code = 404
return response
def handler500(request, *args, **argv):
response = render_to_response('500.html', {},
context_instance=RequestContext(request))
response.status_code = 500
return response
Update
handler404 and handler500 are exported Django string configuration variables found in django/conf/urls/__init__.py. That is why the above config works.
To get the above config to work, you should define the following variables in your urls.py file and point the exported Django variables to the string Python path of where these Django functional views are defined, like so:
# project/urls.py
handler404 = 'my_app.views.handler404'
handler500 = 'my_app.views.handler500'
Update for Django 2.0
Signatures for handler views were changed in Django 2.0: https://docs.djangoproject.com/en/2.0/ref/views/#error-views
If you use views as above, handler404 will fail with message:
"handler404() got an unexpected keyword argument 'exception'"
In such case modify your views like this:
def handler404(request, exception, template_name="404.html"):
response = render_to_response(template_name)
response.status_code = 404
return response