Return Python objects, always get JSON.
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Latest commit 98b1dcd Oct 23, 2016 @jsocol committed on GitHub Merge pull request #33 from Quazer/master
Change default Content-Type value via Django settings



django-jsonview is a simple decorator that translates Python objects to JSON and makes sure your view will always return JSON.

I've copied and pasted this so often I decided I just wanted to put it in a package.


Just install with pip:

pip install django-jsonview

No need to add to INSTALLED_APPS or anything.


Just import the decorator, use, and return a JSON-serializable object:

from jsonview.decorators import json_view

def my_view(request):
    return {
        'foo': 'bar',

Class-based views (CBVs) can either use Django's @method_decorator or can wrap the output of .as_view():

from django.utils.decorators import method_decorator
from jsonview.decorators import json_view

class MyView(View):
    def dispatch(self, *args, **kwargs):
        return super(MyView, self).dispatch(*args, **kwargs)

# or, in URLconf

patterns = [
    url(r'^/my-view/$', json_view(MyView.as_view())),

Content Types

If you need to return a content type other than the standard application/json, you can specify that in the decorator with the content_type argument, for example:

from jsonview.decorators import json_view

def myview(request):
    return {'foo': 'bar'}

The response will have the appropriate content type header.

Return Values

The default case is to serialize your return value and respond with HTTP 200 and a Content-Type of application/json.

The @json_view decorator will handle many exceptions and other cases, including:

  • Http404
  • PermissionDenied
  • HttpResponseNotAllowed (e.g. require_GET, require_POST)
  • jsonview.exceptions.BadRequest (see below)
  • Any other exception (logged to django.request).

Any of these exceptions will return the correct status code (i.e., 404, 403, 405, 400, 500) a Content-Type of application/json, and a response body that looks like:

    'error': STATUS_CODE,
    'message': str(exception),


As of v0.4, application exceptions do not behave this way if DEBUG = False. When DEBUG = False, the message value is always An error occurred. When DEBUG = True, the exception message is sent back.


HTTP does not have a great status code for "you submitted a form that didn't validate," and so Django doesn't support it very well. Most examples just return 200 OK.

Normally, this is fine. But if you're submitting a form via Ajax, it's nice to have a distinct status for "OK" and "Nope." The HTTP 400 Bad Request response is the fallback for issues with a request not-otherwise-specified, so let's do that.

To cause @json_view to return a 400, just raise a jsonview.exceptions.BadRequest with whatever appropriate error message.


If your view raises an exception, @json_view will catch the exception, log it to the normal django.request logger, and return a JSON response with a status of 500 and a body that looks like the exceptions in the Return Values section.


Because the @json_view decorator handles the exception instead of propagating it, any exception middleware will not be called, and any response middleware will be called.

Status Codes

If you need to return a different HTTP status code, just return two values instead of one. The first is your serializable object, the second is the integer status code:

def myview(request):
    if not request.user.is_subscribed():
        # Send a 402 Payment Required status.
        return {'subscribed': False}, 402
    # Send a 200 OK.
    return {'subscribed': True}

Extra Headers

You can add custom headers to the response by returning a tuple of three values: an object, a status code, and a dictionary of headers.

def myview(request):
    return {}, 200, {'X-Server': 'myserver'}

Custom header values may be overwritten by response middleware.

Raw Return Values

To make it possible to cache JSON responses as strings (and because they aren't JSON serializable anyway) if you return an HttpResponse object (or subclass) it will be passed through unchanged, e.g.:

from django import http
from jsonview.decorators import JSON

def caching_view(request):
    kached = cache.get('cache-key')
    if kached:
        return http.HttpResponse(kached, content_type=JSON)
    # Assuming something else populates this cache.
    return {'complicated': 'object'}


@require_POST and the other HTTP method decorators work by returning a response, rather than raising, an an exception, so HttpResponseNotAllowed is handled specially.

Alternative JSON Implementations

There is a healthy collection of JSON parsing and generating libraries out there. By default, it will use the old standby, the stdlib json module. But, if you'd rather use ujson, or cjson or yajl, you should go for it. Just add this to your Django settings:

JSON_MODULE = 'ujson'

Anything, as long as it's a module that has .loads() and .dumps() methods.

Configuring JSON Output

.. versionadded:: 0.5

Additional keyword arguments can be passed to json.dumps() via the JSON_OPTIONS = {} Django setting. For example, to pretty-print JSON output:

    'indent': 4,

Or to compactify it:

    'separators': (',', ':'),

jsonview uses DjangoJSONEncoder by default. To use a different JSON encoder, use the cls option:

    'cls': '',

JSON_OPTIONS['cls'] may be a dotted string or a JSONEncoder class.

.. versionchanged:: 1.0

If you are using a JSON module that does not support the ``cls`` kwarg, such as ujson, set the cls option to None:

    'cls': None,

Default value of content-type is 'application/json'. You can change it vie the JSON_DEFAULT_CONTENT_TYPE Django settings. For example, to add charset:

JSON_DEFAULT_CONTENT_TYPE = 'application/json; charset=utf-8'

Atomic Requests

Because @json_view catches exceptions, the normal Django setting ATOMIC_REQUESTS does not correctly cause a rollback. This can be worked around by explicitly setting @transaction.atomic below the @json_view decorator, e.g.:

def my_func(request):
    # ...


Pull requests and issues welcome! I ask two simple things:

  • Tests, including the new ones you added, must pass. (See below.)
  • The flake8 tool should not return any issues.

Running Tests

To run the tests, you probably want to create a virtualenv, then install Django and Mock with pip:

pip install Django==${DJANGO_VERSION} mock==1.0.1

Then run the tests with:

./ test