RuntimeError: async generator ignored GeneratorExit

[Jonas1312]

Hi,

When I run this code:

gql_ws_transport = WebsocketsTransport(
    url=self.ws_endpoint,
    headers=self.headers,
    subprotocols=[WebsocketsTransport.APOLLO_SUBPROTOCOL],
    init_payload={
        "headers": {"Accept": "application/json", "Content-Type": "application/json"},
        "Authorization": f"X-API-Key: {self.api_key}",
    },
)
gql_ws_client = Client(
    transport=gql_ws_transport, fetch_schema_from_transport=True
)

subscription = gql_ws_client.subscribe(MY_GRAPHQL_QUERY, variables)

for result in subscription:
    should_break = do_something()
    if should_break:
        subscription.close()

I get an error in the console:

Exception ignored in: <async_generator object Client.subscribe_async at 0x7f9858ee77a0>
RuntimeError: async generator ignored GeneratorExit

How can I break the loop without having this error printed out?

Thanks!

System info:

• OS: macos
• Python version: 3.7
• gql version: 3.4.0
• graphql-core version: 3.2.3

[leszekhanusz]

The normal way to exit is simply to use break:

for result in subscription:
    should_break = do_something()
    if should_break:
        break

but the same error appears on version 3.4.0, this is a bug.

I've made the PR #395 which should solve this.

[leszekhanusz]

It should be fixed in the pre-release v3.5.0b3

[Jonas1312]

I tested the pre-release and it seems to have fixed the bug. Thanks a lot!

[NikitaMishin]

It is possible to trigger unsubscription indirectly (not inside the loop)?
This is quite important if the events on the stream are not frequent.

[NikitaMishin]

@leszekhanusz

[leszekhanusz]

@NikitaMishin It should be possible in sync mode by calling subscription.close() from another thread but a better idea would be to use the async mode and run a subscription inside a task. Once you don't need the subscription anymore you can cancel the task. See advanced usage for an example.