不调用Django通道WebSocketConsumer的断开()方法
不调用Django通道WebSocketConsumer的断开()方法
提问于 2020-07-11 09:00:14
我有以下类，它是从
WebSocketConsumer继承的：
from channels.generic.websocket import WebsocketConsumer
from channels.exceptions import StopConsumer
class MyConsumer(WebsocketConsumer):
def connect(self):
self.accept()
def receive(self, text_data=None, bytes_data=None):
data = json.loads(text_data)
n = data["number"]
for i in range(n):
self.send(json.dumps({"number":i}))
def disconnect():
raise StopConsumer输入的JSON只包含一个名为
number的参数。我正在用一个铬插件测试这段代码。当我打开连接并在不发送任何消息的情况下关闭它时，断开连接方法将按预期执行。例如，当数字为100，而接收方法中的循环尚未完成，而我在两者之间断开连接时，将不调用disconnect方法，并得到以下错误：
ERROR - server - Exception inside application: Attempt to send on a closed protocol.
File "MyConsumer.py", line 2, in receive
self.send
File "python3.6/site-packages/channels/generic/websocket.py", line 69, in send
{"type": "websocket.send", "text": text_data},
File "python3.6/site-packages/channels/consumer.py", line 107, in send
self.base_send(message)
File "python3.6/site-packages/asgiref/sync.py", line 64, in __call__
return call_result.result()
File "/usr/local/var/pyenv/versions/3.6.10/lib/python3.6/concurrent/futures/_base.py", line 432, in result
return self.__get_result()
File "/usr/local/var/pyenv/versions/3.6.10/lib/python3.6/concurrent/futures/_base.py", line 384, in __get_result
raise self._exception
File "python3.6/site-packages/asgiref/sync.py", line 78, in main_wrap
result = await self.awaitable(*args, **kwargs)
File "python3.6/site-packages/channels/sessions.py", line 220, in send
return await self.real_send(message)
File "python3.6/site-packages/daphne/server.py", line 198, in handle_reply
protocol.handle_reply(message)
File "python3.6/site-packages/daphne/ws_protocol.py", line 179, in handle_reply
self.serverSend(message["text"], False)
File "site-packages/daphne/ws_protocol.py", line 223, in serverSend
self.sendMessage(content.encode("utf8"), binary)
File "python3.6/site-packages/autobahn/websocket/protocol.py", line 2216, in sendMessage
raise Disconnected("Attempt to send on a closed protocol")
Attempt to send on a closed protocol我希望能够断开连接，即使对给定数字的处理尚未完成。
WebSocketConsumer不可能做到这一点吗？我是否误解了WebSocketConsumer
的同步性质？
Stack Overflow用户
发布于 2020-07-11 09:36:51
当用户连接关闭并继续向it发送数据时，通道将引发异常，请使用try ... except... block
def receive(self, text_data=None, bytes_data=None):
data = json.loads(text_data)
n = data["number"]
try:
for i in range(n):
self.send(json.dumps({"number":i}))
except Exception as e:# i don't remember exact exception, will change this soon
print(e) #you can log this situation or do what ever you want页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/62847184
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