python - 使用 websocket-client 连接时出现 WinError 10057

Question

我正在制作一个 Flask webapp，我正在使用 Flask-Socketio。由于各种原因，我还需要使用 websocket-client 包。一切都按预期工作，除了当我尝试在不同网络上的另一台计算机上运行该应用程序时，我收到以下错误:

"""
Traceback (most recent call last):
  File "[Path to venv]\venv\lib\site-packages\flask\app.py", line 2449, in wsgi_app
    response = self.handle_exception(e)
  File "[Path to venv]\venv\lib\site-packages\flask\app.py", line 1866, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "[Path to venv]\venv\lib\site-packages\flask\_compat.py", line 39, in reraise
    raise value
  File "[Path to venv]\venv\lib\site-packages\flask\app.py", line 2446, in wsgi_app
    response = self.full_dispatch_request()
  File "[Path to venv]\venv\lib\site-packages\flask\app.py", line 1951, in full_dispatch_request
    rv = self.handle_user_exception(e)
  File "[Path to venv]\venv\lib\site-packages\flask\app.py", line 1820, in handle_user_exception
    reraise(exc_type, exc_value, tb)
  File "[Path to venv]\venv\lib\site-packages\flask\_compat.py", line 39, in reraise
    raise value
  File "[Path to venv]\venv\lib\site-packages\flask\app.py", line 1949, in full_dispatch_request
    rv = self.dispatch_request()
  File "[Path to venv]\venv\lib\site-packages\flask\app.py", line 1935, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File "[Path to app]\app\views.py", line 7, in index
    sio.connect("http://localhost:80/", transports=['websocket', 'polling'])
  File "[Path to venv]\venv\lib\site-packages\socketio\client.py", line 262, in connect
    engineio_path=socketio_path)
  File "[Path to venv]\venv\lib\site-packages\engineio\client.py", line 170, in connect
    url, headers, engineio_path)
  File "[Path to venv]\venv\lib\site-packages\engineio\client.py", line 346, in _connect_websocket
    cookie=cookies)
  File "[Path to venv]\venv\lib\site-packages\websocket\_core.py", line 514, in create_connection
    websock.connect(url, **options)
  File "[Path to venv]\venv\lib\site-packages\websocket\_core.py", line 223, in connect
    options.pop('socket', None))
  File "[Path to venv]\venv\lib\site-packages\websocket\_http.py", line 120, in connect
    sock = _open_socket(addrinfo_list, options.sockopt, options.timeout)
  File "[Path to venv]\venv\lib\site-packages\websocket\_http.py", line 189, in _open_socket
    raise error
  File "[Path to venv]\venv\lib\site-packages\websocket\_http.py", line 172, in _open_socket
    sock.connect(address)
OSError: [WinError 10057] A request to send or receive data was disallowed because 
the socket is not connected and (when sending on a datagram socket using a sendto call) 
no address was supplied
"""

我已经尽可能地将我的代码归结为以下内容，它仍然可以在我的计算机上运行，​​但在另一个上给出了相同的错误:

|start.py
|app
    |__init__.py
    |views.py
    |templates
        |index.html

# __init__.py

from flask import Flask
from flask_socketio import SocketIO
from gevent import monkey

monkey.patch_all()

APP = Flask(__name__)
SOCKETIO = SocketIO(APP, engineio_logger=True, logger=True)

from . import views

# views.py

from app import APP
from socketio import Client
from flask import render_template

@APP.route('/', methods=['GET', 'POST'])
def index():
    sio = Client()
    sio.connect("http://localhost:80", transports=['websocket', 'polling']) # Error being caused here
    return render_template('index.html')

# start.py

from app import APP, SOCKETIO

if __name__ == "__main__":
    SOCKETIO.run(APP, debug=True, port=80, host='0.0.0.0')

index.html只是一个基本的“Hello World”html 页面。

什么样的东西可能在一台计算机/网络上而不是另一台计算机/网络上给我这个错误，尤其是在 localhost:80 上运行它?我真的不知道在这里尝试什么。

编辑:将回溯数据添加到错误

编辑 2:在我的实际代码中，websocket.Client 正在 Celery 任务中运行。我没有在此处包含它，因为该错误是可以重现的，而无需太复杂。

Answer 1

问题是当你尝试运行 -(Flask-webapp, websocket-client)- both of them同时，only one其中将运行。

更新:

以下是一些需要注意的事项:

如果您更换机器，请确保系统软件已经是最新的。

检查您的系统是否位于防火墙后面，但是当您在同一台机器上运行 Flask webapp 和 websocket-client 时，则不需要此步骤。

更新操作系统后尝试重新启动计算机。

使用multiprocessing而不是 Threads因为他们不会在这里工作，因为著名的 Python-Global-Interpreter or GIL .根据Only one thread can be in a state of execution at any point in time.执行 GIL 的开发人员看不到 GIL 的影响
单线程程序，但它可以是 - 性能瓶颈 - 在
CPU 绑定(bind)和多线程代码。

然后检查:How to run flask on Port:80 . For development purposes端口:5000 is mostly recommended.

为了执行"Error Handling"访问flask-socketio文档，有很好的提示来处理异常。

您可以按照以下方式进行:

@socketio.on_error_default  # handles all namespaces without an explicit error handler
def default_error_handler(e):
    pass

最后，处理 socket.timeout ，您也可以使用以下代码来处理最终错误:套接字超时:

try:
socketio.run(app,...

除了socket.error作为socketerror:
print("错误: ", socketerror)

使用 postman 会更好。 .对于learning and testing目的，尝试使用POSTMAN这是一个发送REST的快速简便的平台, SOAP , 和 GraphQL直接在 Postman 中请求。访问引用:getting-started-postman .

这是一个示例程序:

import multiprocessing

# Assume for flask-app
def start_flask_app:
    app.run()

# Assume for websocket-client-app
# Assume basic events you will handle in your client e.g. on_message, on_error, on_close
def start_ws:
    ws = websocket.WebSocketApp(WS_URI, on_message= on_message,
                                        on_error = on_error, 
                                        on_close = on_close)
    ws.on_open = on_open
    ws.run_forever()

# use of sub-processes instead of threads.
p_flask = multiprocessing.Process(target=start_flask_app)
p_ws = multiprocessing.Process(target=start_ws)

p_ws.start()
p_flask.start()

p_ws.join()
p_flask.join()