WebMar 12, 2024 · 1 Answer. For a consumer to receive a message, that consumer needs to be plugged into a Router. The most common case is to use a URLRouter where the Consumer will receive events from a connection (for example a web socket). You can use a ChannelNameRouter to create a consumer that will listen on a specific Chanel. WebAug 7, 2024 · application = ProtocolTypeRouter({ "channel": ChannelNameRouter({ "testing-print": PrintConsumer, }), }) consumers.py. from channels.consumer import …

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WebJul 4, 2024 · # routing.py from channels.routing import ProtocolTypeRouter, ChannelNameRouter from.consumers import SlackConsumer application = ProtocolTypeRouter ({'channel': ChannelNameRouter ({'slack': SlackConsumer,}),}) Start a Channels worker to handle Slack events from the channel layer. python manage.py … Web"channel": ChannelNameRouter({ 'o-ready': Consumer, 'o-progress': ProgressConsumer, }) }) I run two workers: one for “o-ready” channel and one for “o-progress”. The issue is when the ProgressConsumer is instantiated and receiving a message the self.channel_layer points to “default” layer instead of “progress” layer. harvel whippet racing club

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WebWorker and Background Tasks¶. While channel layers are primarily designed for communicating between different instances of ASGI applications, they can also be used … Webdjango - 如何使用 ChannelNameRouter 在 Worker 和 Websocket(Django 和 Channels2.x)之间进行通信？ 标签 django python-3.x websocket django-channels. 我正在尝试设置一个 … WebMar 14, 2024 · from django.urls import path from channels.routing import ChannelNameRouter, ProtocolTypeRouter, URLRouter from rest_framework_reactive.consumers import ClientConsumer, MainConsumer, WorkerConsumer from rest_framework_reactive.protocol import CHANNEL_MAIN, … harvend learn programing