Django频道websocket重新连接

问题描述 投票:0回答:1

我正在Django中构建一个消息传递系统,使用网络套接字来接收消息。

这是我的尝试,

from channels.consumer import AsyncConsumer
from channels.db import database_sync_to_async

class ChatConsumer(AsyncConsumer):

    async def websocket_connect(self, event):
        print("connected", event)
        await self.send({
            "type": "websocket.accept"
        })

    async def websocket_receive(self, event):
        print("receive", event)
        message = event['text']
        obj = await self.create_obj(message=message)
        print('Created...')

    async def websocket_disconnect(self, event):
        print("disconnected", event)

    @database_sync_to_async
    def create_obj(self, **kwargs):
        obj = ChatMessage.objects.create(message=kwargs['message'])
        return obj

当我启动客户端应用程序时,Web套接字已连接,并且我可以接收到消息并且可以存储到数据库中。

connected {'type': 'websocket.connect'}
WebSocket CONNECT /ws/message [192.168.0.101:50148]

经过一些空闲时间后,网络套接字会自动断开,

Application instance <Task pending coro=<SessionMiddlewareInstance.__call__() running at /Users/ajay/Desktop/workspace/projects/python/django/websock/venv/lib/python3.7/site-packages/channels/sessions.py:183> wait_for=<Future pending cb=[<TaskWakeupMethWrapper object at 0x10b311510>()]>> for connection <WebSocketProtocol client=['192.168.0.101', 50148] path=b'/ws/message'> took too long to shut down and was killed.

断开连接后,我无法接收消息,当我重新加载客户端时,Web-socket已连接,有什么方法可以自动重新连接we-socket而无需重新加载客户端。我要求引导我一些建议以实现这一目标,这对我非常有帮助,谢谢。

django websocket django-channels reconnect
1个回答
0
投票

似乎您的客户端终止了连接。当用作客户端导入websocket时,我面临相同的问题。

 # in main while loop
 while not self.finished:
        try:
            self.websocket = await websockets.client.connect(
                self.connect_url,
                max_size=None,
                extra_headers=self.extra_headers
            )
.....
# in your`s producer
.......
 while True:
        # Make sure connection is still live.
        pong_waiter = await self.websocket.ping()
        await pong_waiter # if you will not await in 10-20 sec connection will close.
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