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