Question

我正在尝试编写一个websocket，它连接到在localhost上运行的服务，但它正在抛出错误

>>> from websocket import create_connection
>>> ws = create_connection("ws://127.0.0.1", http_proxy_port="2974", http_proxy_host="quividicontent")

Traceback (most recent call last):
  File "<pyshell#24>", line 1, in <module>
    ws = create_connection("ws://127.0.0.1", http_proxy_port="2974", http_proxy_host="quividicontent")
  File "C:\Python27\lib\websocket\_core.py", line 487, in create_connection
    websock.connect(url, **options)
  File "C:\Python27\lib\websocket\_core.py", line 211, in connect
    options.pop('socket', None))
  File "C:\Python27\lib\websocket\_http.py", line 64, in connect
    hostname, port, is_secure, proxy)
  File "C:\Python27\lib\websocket\_http.py", line 97, in _get_addrinfo_list
    addrinfo_list = socket.getaddrinfo(phost, pport, 0, 0, socket.SOL_TCP)
gaierror: [Errno 11001] getaddrinfo failed
>>> ws = create_connection("ws://127.0.0.1", http_proxy_port="2974")

Traceback (most recent call last):
  File "<pyshell#25>", line 1, in <module>
    ws = create_connection("ws://127.0.0.1", http_proxy_port="2974")
  File "C:\Python27\lib\websocket\_core.py", line 487, in create_connection
    websock.connect(url, **options)
  File "C:\Python27\lib\websocket\_core.py", line 214, in connect
    self.handshake_response = handshake(self.sock, *addrs, **options)
  File "C:\Python27\lib\websocket\_handshake.py", line 65, in handshake
    status, resp = _get_resp_headers(sock)
  File "C:\Python27\lib\websocket\_handshake.py", line 122, in _get_resp_headers
    raise WebSocketBadStatusException("Handshake status %d", status)
WebSocketBadStatusException: Handshake status 200
>>> import socket
>>> socket.getaddrinfo('localhost', 2974)
[(23, 0, 0, '', ('::1', 2974, 0, 0)), (2, 0, 0, '', ('127.0.0.1', 2974))]

我确信端口是开放式的，正在使用套接字以及onOpen onMessage函数的更复杂示例。

Answer 1

出现此问题是因为您需要为websocket指定握手协议。

来自wikipedia：

要建立WebSocket连接，客户端会发送WebSocket 握手请求，服务器返回WebSocket握手响应，如下例所示。

在此握手过程中，服务器和客户端讨论应该使用哪些协议，如果没有指定协议，这可能会导致错误（并非总是如此）。

从python-websocket doc中，指定一个可以完成的协议：

ws = websocket.create_connection("ws://exapmle.com/websocket", subprotocols=["binary", "base64"])

在您的特定情况下，subprotocols应为['quividicontent']

Python websockets错误gaierror

1 个答案: