Django Channels Example issues

[jonatasbaldin]

• Python version: 3.6
• Operating System: OSX Sierra

Description

When testing the Django Channels examples, I went through some errors:

1: graphql_ws/django_channels.py file is not on the latest package version. But I downloaded it manually inside the package.

2: I was getting the error "Subscriptions are not allowed. You will need to either use the subscribe function or pass allow_subscriptions=True". And, since the GraphQLView object does not support the allow_subscriptions parameter, I changed it inside the package to True (just for testing).

3. After that, I was getting the error Subscription must return Async Iterable or Observable. Reading this issue, I decided to add 'MIDDLEWARE: [] into the GRAPHENE settings variable.

4. After that, I started receiving this error: 'AnonymousObservable' object has no attribute 'errors'. Then I got a little bit frustrated and stop trying 😅

Does anyone have a clue why this is happening?

Thanks!

[nick-lehmann]

I have experienced exactly the same today, but after seeing your issue I have stopped trying already after the second step.

[jonatasbaldin]

I'm whiling to help, but I don't understand too much about the Observables :(

[mlugowska]

I've checked that DjangoDebugMiddleware returns promise, so it cannot be set together with subscriptions and observables. But AnonymousObservable' object has no attribute 'errors still occures even when I added field errors to observable. I think that it's graphene bug (?). But I still really need to add subscriptions to my django app (graphene-django, python). Do you have any thoughts how can I do this? I know graphene-django-subscriptions lib, but does it really necessery to use serializer class only in case of subscriptions? Please, a little help will be desirable.

[colanconnon]

@jonatasbaldin which version of channels are you using?

[fcalo]

Changing the urls.py it's solved
eamigo86/graphene-django-subscriptions#2

[xDHILEx]

I'm stuck on the AnonymousObserable issue as well. https://github.com/xDHILEx/graphqlwsdjangochannels. @colanconnon I'm using channels 1.1.8

[colanconnon]

@xDHILEx I will have to look at this closer when I have some time

[adrianmoisey]

Has anyone managed to solve this? I'm getting the same issue.

[ambientlight]

just discovered graphene and though of evaluating state of graphql integration with django.
this might a bit late though:

For short:

1. To resolve Subscriptions are not allowed. You will need to either use the subscribe function or pass allow_subscriptions=True after navigating the call stack I found we can pass a custom backend to GraphQLView that can custom exc_context parameters.

from graphql.backend import GraphQLCoreBackend
class GraphQLCustomCoreBackend(GraphQLCoreBackend):
    def __init__(self, executor=None):
        # type: (Optional[Any]) -> None
        super().__init__(executor)
        self.execute_params['allow_subscriptions'] = True

2. Same way, used MIDDLEWARE: [] in GRAPHENE settings. as OP.
3. To resolve 'AnonymousObservable' object has no attribute 'errors'. a custom view can to be defined that would unbox execution_result from the Observable. (this is hacky) Using ExtraGraphQLView, AuthenticatedGraphQLView outlined in 'AnonymousObservable' object has no attribute 'errors' eamigo86/graphene-django-subscriptions#2 didn't work for me as they just as GraphQLView don't handle observables returned from execute_graphql_request which is from graphql/execution/executor.py#L265

class GraphQLObservableUnboxingView(GraphQLView):
    def execute_graphql_request(
            self, request, data, query, variables, operation_name, show_graphiql=False
    ):
        target_result = None

        def override_target_result(value):
            nonlocal target_result
            target_result = value

        execution_result = super().execute_graphql_request(request, data, query, variables, operation_name, show_graphiql)
        if execution_result:
            if isinstance(execution_result, ObservableBase):
                target = execution_result.subscribe(on_next=lambda value: override_target_result(value))
                target.dispose()
            else:
                return execution_result

        return target_result

So in urls.py it can be:

url(r'^graphql', GraphQLObservableUnboxingView.as_view(graphiql=True, backend=GraphQLCustomCoreBackend()))

I have not used django channels and just tried reproducing the most possible minimal example so that subscription can be resolved:

subscription{
  subscribeToFoo(id: 1)
}

from rx import Observable
class Subscription(graphene.ObjectType):
    subscribe_to_foo = graphene.Boolean(id=graphene.Int())
    
    def resolve_subscribe_to_foo(self, args, **kwargs):
        return Observable.of(True)

schema = graphene.Schema(query=Query, subscription=Subscription)

also please make sure you are using rxpy 1.6.*

[hoffme]

what is it ObservableBase?

[carlosalvarez91]

@ambientlight your solution seems to work when MIDDLEWARE: [ ], but since I'm using 'graphql_jwt.middleware.JSONWebTokenMiddleware', the rest of queries and mutations throws a Not logged in Error...

[ambientlight]

@phjocoronel2806: ObservableBase is from rxpy: https://github.com/ReactiveX/RxPY/blob/release/v1.6.x/rx/core/observablebase.py

@carlosalvarez91: haven't used any middleware with the above test so cannot yet give you any feedback on this.