Common Tastypie-django error handling

Question

I would like to return JSON responses back instead of just returning a header with an error code. Is there a way in tastypie to handle such errors?

Answer 1

Thought about it in the end. This is a good resource to look at if someone else needs it. http://gist.github.com/1116962

class YourResource(ModelResource): def wrap_view(self, view): """ Wraps views to return custom error codes instead of generic 500's """ @csrf_exempt def wrapper(request, *args, **kwargs): try: callback = getattr(self, view) response = callback(request, *args, **kwargs) if request.is_ajax(): patch_cache_control(response, no_cache=True) # response is a HttpResponse object, so follow Django instructions # to change it to your needs before you return it. # https://docs.djangoproject.com/en/dev/ref/request-response/ return response except (BadRequest, ApiFieldError), e: return HttpBadRequest({'code': 666, 'message':e.args[0]}) except ValidationError, e: # Or do some JSON wrapping around the standard 500 return HttpBadRequest({'code': 777, 'message':', '.join(e.messages)}) except Exception, e: # Rather than re-raising, we're going to things similar to # what Django does. The difference is returning a serialized # error message. return self._handle_500(request, e) return wrapper 

Answer 2

You can overwrite the tastypie Resource _handle_500() method. The fact that it starts with an underscore does indicate that it is a "private" method and should not be overwritten, but I find it cleaner than overwriting wrap_view() and replicating a lot of logic.

Here is how I use it:

 from tastypie import http from tastypie.resources import ModelResource from tastypie.exceptions import TastypieError class MyResource(ModelResource): class Meta: queryset = MyModel.objects.all() fields = ('my', 'fields') def _handle_500(self, request, exception): if isinstance(exception, TastypieError): data = { 'error_message': getattr( settings, 'TASTYPIE_CANNED_ERROR', 'Sorry, this request could not be processed.' ), } return self.error_response( request, data, response_class=http.HttpApplicationError ) else: return super(MyResource, self)._handle_500(request, exception) 

In this case, I catch all Tastypie errors by checking if the exception instance of TastypieError and returns a JSON response with the message "Sorry, this request could not be processed." If this is another exception, I call the parent _handle_500 with super() , which will create a django error page in development mode or send_admins() in send_admins() mode.

If you want to have a specific JSON response for a specific exception, just do an isinstance() check for the specific exception. Here are all the Tastypie exceptions:

https://github.com/toastdriven/django-tastypie/blob/master/tastypie/exceptions.py

Actually, I think there should be a better / cleaner way to do this in Tastypie, so I opened a ticket on my github.