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Home/ Questions/Q 460279
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T22:51:27+00:00 2026-05-12T22:51:27+00:00

I’m looking for an excuse to learn Django for a new project that has

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I’m looking for an excuse to learn Django for a new project that has come up. Typically I like to build RESTful server-side interfaces where a URL maps to resources that spits out data in some platform independent context, such as XML or JSON. This is
rather straightforward to do without the use of frameworks, but some of them such as Ruby on Rails conveniently allow you to easily spit back XML to a client based on the type of URL you pass it, based on your existing model code.

My question is, does something like Django have support for this? I’ve googled and found some ‘RESTful’ 3rd party code that can go on top of Django. Not sure if I’m too keen on that.

If not Django, any other Python framework that’s already built with this in mind so I do not have to reinvent the wheel as I already have in languages like PHP?

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  1. Editorial Team
    Editorial Team
    2026-05-12T22:51:28+00:00Added an answer on May 12, 2026 at 10:51 pm

    This is probably pretty easy to do.

    URL mappings are easy to construct, for example:

    urlpatterns = patterns('books.views',
      (r'^books/$', 'index'),
      (r'^books/(\d+)/$', 'get'))
    

    Django supports model serialization, so it’s easy to turn models into XML:

    from django.core import serializers
    from models import Book
    
    data = serializers.serialize("xml", Book.objects.all())
    

    Combine the two with decorators and you can build fast, quick handlers:

    from django.http import HttpResponse
    from django.shortcuts import get_object_or_404
    
    def xml_view(func):
      def wrapper(*args, **kwargs):
        result = func(*args, **kwargs)
        return HttpResponse(serializers.serialize("xml", result),
            mimetype="text/xml")
      return wrapper
    
    @xml_view
    def index(request):
      return Books.objects.all()
    
    @xml_view
    def get(request, id):
      return get_object_or_404(Book, pk=id)
    
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