Allow only one message to be in queue

[theobouwman]

Is it possible to allow only one message to be in queue? Such that we you keep publishing messages the current message gets overwritten?

We have a problem where our worker service could crash (i know we need to fix this). But what happens next is that messages keep piling up if we keep publishing every 5 minutes. Resulting in the worker not being able to easily spin up and process other messages.

How we publish:

await broker.publish(CacheDiscoverRefreshOrganisation(organisation_id=organisation_id), list="cache.discover.refresh.organisation")

Our subscriber:

@broker.subscriber(
    list="cache.discover.refresh.organisation",
)
@inject
async def handler(
    msg: CacheDiscoverRefreshOrganisation,
    discover_service: Annotated[DiscoverService, FromDishka()],
):

[Lancetnik]

@theobouwman hi! Why u can't just impl this logic on publisher side? "Do not publish if message present" I mean

[theobouwman]

Is it possible wirth FastStream to check this?

[Lancetnik]

You can use raw redis client from broker to use any underly methods

from faststream.redis import Redis

@broker.subscriber(...)
async def handler(
   redis: Redis  # get connection from context
): ...