Subscriptions are not allowed.

[masarliev]

I followed the example and I got
Subscriptions are not allowed. You will need to either use the subscribe function or pass allow_subscriptions=True
What am I missing?

[eamigo86]

Hi @masarliev, you must make sure that the graphql url of your project defines it using the view graphene_django_extras.views.ExtraGraphQLView, something like that

 from graphene_django_extras.views import ExtraGraphQLView

urlpatterns = [
# .....
 url(r'^graphql', ExtraGraphQLView.as_view(graphiql=True, schema=schema), name='graphql'),
]

Please can you please let me know if the error was corrected. Right now I am working on an update of this module and its documentation

[masarliev]

Thanks for the fast replay. I am still getting the same error. Later today or tomorrow I can provide simple application

[masarliev]

graphql-python/graphql-ws#8
I have the same problem as here

[masarliev]

https://github.com/masarliev/subs-example
@eamigo86 here is the example project

[eamigo86]

Hi @masarliev , I was reviewing your project and apparently it lacks the definition of the file asgi.py, and it's mandatory for Channles.
Could you send me what the command "pip freeze" returns?

[masarliev]

for this I was using docker python:3.6 image and install all required packages from requirements.txt
I found that using python setup.py install from master branch works like charm

[ambientlight]

defining a custom GraphQLCoreBackend can pass allow_subscriptions to execution context like:

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

then it can be passed to your GraphQL view (say):

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

graphql-python/graphql-ws#8 (comment)