Django REST framework: non-model serializer

Question

I am beginner in Django REST framework and need your advice. I am developing a web service. The service has to provide REST interface to other services. The REST interface,

Answer 1

In the urls.py, the function login_required requires

from django.contrib.auth.decorators import login_required

Answer 2

Django-rest-framework works well even without tying it to a model. Your approach sounds ok, but I believe you can trim some of the steps to get everything working.

For example, rest framework comes with a few built-in renderers. Out of the box it can return JSON and XML to the API consumer. You can also enable YAML by just installing the required python module. Django-rest-framework will output any basic object like dict, list and tuple without any extra work on your part.

So basically you only have to create the function or class that takes in arguments, does all of the required calculations and returns its results in a tuple to the REST api view. If JSON and/or XML fits your needs, django-rest-framework will take care of the serialization for you.

You can skip steps 2 and 3 in this case, and just use one class for calculations and one for presentation to the API consumer.

Here are a few snippets may help you out:

Please note that I have not tested this. It's only meant as an example, but it should work :)

The CalcClass:

class CalcClass(object):

    def __init__(self, *args, **kw):
        # Initialize any variables you need from the input you get
        pass

    def do_work(self):
        # Do some calculations here
        # returns a tuple ((1,2,3, ), (4,5,6,))
        result = ((1,2,3, ), (4,5,6,)) # final result
        return result

The REST view:

from rest_framework.views import APIView
from rest_framework.response import Response
from rest_framework import status

from MyProject.MyApp import CalcClass


class MyRESTView(APIView):

    def get(self, request, *args, **kw):
        # Process any get params that you may need
        # If you don't need to process get params,
        # you can skip this part
        get_arg1 = request.GET.get('arg1', None)
        get_arg2 = request.GET.get('arg2', None)

        # Any URL parameters get passed in **kw
        myClass = CalcClass(get_arg1, get_arg2, *args, **kw)
        result = myClass.do_work()
        response = Response(result, status=status.HTTP_200_OK)
        return response

Your urls.py:

from MyProject.MyApp.views import MyRESTView
from django.conf.urls.defaults import *

urlpatterns = patterns('',
    # this URL passes resource_id in **kw to MyRESTView
    url(r'^api/v1.0/resource/(?P<resource_id>\d+)[/]?$', login_required(MyRESTView.as_view()), name='my_rest_view'),
    url(r'^api/v1.0/resource[/]?$', login_required(MyRESTView.as_view()), name='my_rest_view'),
)

This code should output a list of lists when you access http://example.com/api/v1.0/resource/?format=json. If using a suffix, you can substitute ?format=json with .json. You may also specify the encoding you wish to get back by adding "Content-type" or "Accept" to the headers.

[
  [
    1, 
    2, 
    3
  ], 
  [
    4, 
    5, 
    6
  ]
]

Hope this helps you out.