use angular $http with django urlpatterns to get a json file

Question

what I am trying to do is send a $http GET request to a django url pattern and return a .json file. Can I use urlpatterns to return a file instead of a view?

is this even possible?

currently form/data is returning 500

testModule.js

angular.module('mod1', ['mod2'])

.controller('resorcesCtrl', ['mod2Factory',
     function(mod2Factory) {
        mod2Factory.getJSON();
     }
]);


angular.module('mod2', [])
   .factory('mod2Factory', [ '$http',
      function($http) {

        var getJSON = function() {
           return $http({
           url: "/form/data",
           method: "GET"
        })
      }

      return {
         getJSON:getJSON
      }; 
    }
  ])

urls.py

url(r'^form/data', 'myApp.views.getJSON', name='getJSON'),

views.py

def getJSON(request):
    context = {
        "json": json
    }
    return render(request, '', context)

What is the content of your 500 error? Can you put it here with the stacktrace? Which version of Django? — Fernando Macedo
– Fernando Macedo, Commented Mar 24, 2015 at 23:17

Fernando Macedo · Accepted Answer · 2015-03-24 23:30:40Z

2

Starting with Django 1.7, there is a new response object called JsonResponse.

Your view can be as simple as:

from django.http import JsonResponse

def getJSON(request):
    # context is a Python dict.
    context = {
        "key": "value",
        "other_key": 365,
    }
    return JsonResponse(context)

The context must be a valid JSON serializable object. See the docs for more detail about JsonResponse. If you have any trouble with serialization, you can pass your own encoder to JsonResponse to handle custom data types, read more.

edited Mar 24, 2015 at 23:30

answered Mar 24, 2015 at 23:24

Fernando Macedo

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Comments

adrien · Accepted Answer · 2015-03-25 09:25:58Z

1

I prefer use a function like that

import json
from django.http import HttpResponse

def json_responder(**kwargs):
    return HttpResponse(json.dumps(kwargs), content_type="application/json")

def yourview(request):
    return json_responder(type='error', msg='a message")

and that can be used on every django version

answered Mar 25, 2015 at 9:25

adrien

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