python - Django Channels WebSocketConsumers 是无状态的吗

Question

我正在尝试使用 Django channel 创建一个对象，该对象对于连接到套接字的每个人都保持持久性/

当我尝试创建一个在多次 receive() 运行之间保持持久状态的对象时，它会抛出 NoneType 异常

class MyConsumer(WebsocketConsumer):


    def __init__(self,path):
        self.protocol = None
        WebsocketConsumer.__init__(self, path)


    def connection_groups(self):
        return ["test"]

    # Connected to websocket.connect
    def connect(self,message):
        try:
            self.protocol = "hello"
        except Exception as exc:
            print ("Unable to accept incoming connection.  Reason: %s" % str(exc))
        self.message.reply_channel.send({"accept": True})


    # Connected to websocket.receive
    def receive(self,text=None, bytes=None):
        text = self.protocol[1] # This throws an error that says protocol is none
        self.send(text=text, bytes=bytes)

    # Connected to websocket.disconnect
    def disconnect(self,message):
        pass

Answer 1

基于类的消费者是未实例化的，即。 e.每次将新消息路由到消费者时，它都会创建一个全新的消费者。因此，无法在消费者本身的消息之间持久保存数据。