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

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

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:

json.dumps({
    'error': STATUS_CODE,
    'message': str(exception),
})

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.

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:

@json_view
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}

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.

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

Custom header values may be overwritten by response middleware.

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, it.

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.

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:

./run.sh test