Re: Django url pass through

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Hi Robert,

In those cases a HTTP 404 is the right response, to help Search Engines realise that such a URL doesn't exist and to show users a "Page Not Found" message. To create a nice looking 404 response for users just write a view to handle it and add the "handler404" entry to your root URLConf module (the one near the settings.py module).

Adding something like this to urls.py:

handler404 = 'views.http404_handler'

And creating that view in the corresponding views.py module:

from django.http import HttpResponseNotFound

from django.template import loader, RequestContext

def http404_handler(request):

    t = loader.get_template("404.html")

    return HttpResponseNotFound(

        t.render(RequestContext(request, {'request_path': request.path})))

Also create the template 404.html in your templates directory.

Read more on the topic here: https://docs.djangoproject.com/en/1.7/ref/views/#the-404-page-not-found-view

Good luck,

Daniel

On 23 Sep 2014, at 04:25, robert brook <software.by.python@gmail.com> wrote:

How is a url conf written so that if none of the useful urls are matched it will pass through to some sort of wild card regular expressions so that the view / redirection can be performed gracefully

Thanks

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