django-jsonview is a simple decorator that translates Python objects to JSON and makes sure your view will always return JSON.
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',
}
Class-based views (CBVs) can inherit from JsonView, use Django's
@method_decorator
or wrap the output of .as_view()
# inherit from JsonView
from jsonview.views import JsonView
class MyView(JsonView):
def get_context_data(self, **kwargs):
context = super(MyView, self).get_context_data(**kwargs)
context['my_key'] = 'some value'
return context
# or, method decorator
from django.utils.decorators import method_decorator
from jsonview.decorators import json_view
class MyView(View):
@method_decorator(json_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())),
]
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
@json_view(content_type='application/vnd.github+json')
def myview(request):
return {'foo': 'bar'}
The response will have the appropriate content type header.
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),
})
Note
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.
Note
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.
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.
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
@json_view
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'}
Note
@require_POST
and the other HTTP method decorators work by
returning a response, rather than raising, an exception, so
HttpResponseNotAllowed
is handled specially.
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.
Additional keyword arguments can be passed to json.dumps()
via the
JSON_OPTIONS = {}
Django setting. For example, to pretty-print JSON
output
JSON_OPTIONS = {
'indent': 4,
}
Or to compactify it
JSON_OPTIONS = {
'separators': (',', ':'),
}
jsonview uses DjangoJSONEncoder
by default. To use a different JSON
encoder, use the cls
option
JSON_OPTIONS = {
'cls': 'path.to.MyJSONEncoder',
}
JSON_OPTIONS['cls']
may be a dotted string or a JSONEncoder
class.
If you are using a JSON module that does not support the ``cls``
kwarg, such as ujson, set the cls
option to None
JSON_OPTIONS = {
'cls': None,
}
Default value of content-type is 'application/json'. You can change it
via the JSON_DEFAULT_CONTENT_TYPE
Django settings. For example, to
add charset
JSON_DEFAULT_CONTENT_TYPE = 'application/json; charset=utf-8'
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.
@json_view
@transaction.atomic
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.)
- Coverage should not drop below 100. You can install
coverage
with pip and run./run.sh coverage
to check. - 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