从视图访问消费者，通过WebSocket(Django-Channelsv2.2和DRF)将api接收的数据推送给用户

Question

我发现自己很难找到一种方法将数据从View(APIView Django-rest-framework(或分布式任务(Celery/RabbitMQ直接(推送到消费者(Django Channels v.2.2(

我

真的很感激任何允许我从 View 访问使用者的代码示例，因为触发器本身是输入设备，而不是 Web，因此我可以从它进行 API 调用，所以我需要连接器。

Django Channels v.2.2，DjangoChannelsRestFramework，主要是这里描述的所有情况：https://channels.readthedocs.io/en/latest/community.html 以及相关问题@stackoverflow

consumers.py

class BasicConsumer(AsyncConsumer):
    async def websocket_connect(self, event):
        print('connected', event)
        await self.send({
            "type": "websocket.accept"
        })
        # await self.send({
        #     "type": "websocket.send",
        #     "text": "Hello world"
        # })
    async def websocket_receive(self, event):
        print('receive', event)
        front_text = event.get('text', None)
        if front_text is not None:
            loaded_dict_data = json.loads(front_text)
            msg = loaded_dict_data.get('message')
            print(msg)
            user = self.scope['user']
            if user.is_authenticated:
                username = user.username
            myResponse = {
                'message': msg,
                'username': username
            }
            await self.send({
                "type": "websocket.send",
                "text": json.dumps(myResponse)
            })
    async def websocket_disconnect(self, event):
        print('disconnect', event)
    @database_sync_to_async
    def get_thread(self, user, other_username):
        return Thread.objects.get_or_new(user, other_username)[0]

views.py

class BasicView(APIView):
    def post(self, request):
        serializer = BasicViewSerializer(data=request.data)
        if serializer.is_valid():
           triggerConsumer(serializer.validated_data)

Answer 1

我认为最好

定义一个通道名称并向它发送数据。

from asgiref.sync import async_to_sync
async_to_sync(channel_layer.send)("channel_name", {...})

您可以在此处查看完整文档 文档