我正在尝试使用RFC 6455协议在python上实现一个简单的websoket服务器。 我从here和here获取了握手格式。
我使用Chromium 17和Firefox 11作为客户端,并收到此错误:
Uncaught Error: INVALID_STATE_ERR: DOM Exception 11
我希望在浏览器中看到hello from server
,在服务器日志中看到hello from client
。
我猜我的握手是错的,你能指出我的错误吗?
GET / HTTP/1.1
Upgrade: websocket
Connection: Upgrade
Host: 127.0.0.1:8999
Origin: null
Sec-WebSocket-Key: 8rYWWxsBPEigeGKDRNOndg==
Sec-WebSocket-Version: 13
HTTP/1.1 101 Switching Protocols
Upgrade: websocket
Connection: Upgrade
Sec-WebSocket-Accept: 3aDXXmPbE5e9i08zb9mygfPlCVw=
HTTP/1.1 101 Switching Protocols\r\nUpgrade: websocket\r\nConnection: Upgrade\r\nSec-WebSocket-Accept: 3aDXXmPbE5e9i08zb9mygfPlCVw=\r\n\r\n
import socket
import re
from base64 import b64encode
from hashlib import sha1
websocket_answer = (
'HTTP/1.1 101 Switching Protocols',
'Upgrade: websocket',
'Connection: Upgrade',
'Sec-WebSocket-Accept: {key}\r\n\r\n',
)
GUID = "258EAFA5-E914-47DA-95CA-C5AB0DC85B11"
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind(('127.0.0.1', 8999))
s.listen(1)
client, address = s.accept()
text = client.recv(1024)
print text
key = (re.search('Sec-WebSocket-Key:\s+(.*?)[\n\r]+', text)
.groups()[0]
.strip())
response_key = b64encode(sha1(key + GUID).digest())
response = '\r\n'.join(websocket_answer).format(key=response_key)
print response
client.send(response)
print client.recv(1024)
client.send('hello from server')
<!DOCTYPE html>
<html>
<head>
<title>test</title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<script type="text/javascript">
var s = new WebSocket('ws://127.0.0.1:8999');
s.onmessage = function(t){alert(t)};
s.send('hello from client');
</script>
</head>
<body>
</body>
</html>
答案 0 :(得分:3)
您的服务器握手代码看起来不错。
客户端代码看起来会在(异步)握手完成之前尝试发送消息。您可以通过将消息发送到websocket的onopen方法来避免这种情况。
建立连接后,服务器不会以纯文本形式发送或接收消息。有关详细信息,请参阅规范的data framing部分。 (客户端代码可以忽略这一点,因为浏览器会为您处理数据框架。)