Twisted websocket chatserver openid身份验证

Question

我有一个python聊天服务器，使用twisted和autobahn websockets连接。

factory = MessageServerFactory("ws://localhost:9000", debug=debug, debugCodePaths=debug)
factory.protocol = MessageServerProtocol
factory.setProtocolOptions(allowHixie76=True)
listenWS(factory)

这是服务器

import logging
from autobahn.websocket import WebSocketServerFactory, WebSocketServerProtocol
from DatabaseConnector import DbConnector
from LoginManager import LoginManager
from MessageTypes import MessageParser
class MessageServerProtocol(WebSocketServerProtocol):
def onOpen(self):
    self.factory.register(self)
def onMessage(self, msg, binary):
    if not binary:
        self.factory.processMessage(self, msg)
def connectionLost(self, reason):
    WebSocketServerProtocol.connectionLost(self, reason)
    self.factory.unregister(self)

class MessageServerFactory(WebSocketServerFactory):
logging.basicConfig(filename='log/dastan.log',format='%(levelname)s:%(message)s',level=logging.WARNING)
def __init__(self, url, debug=False, debugCodePaths=False):
    WebSocketServerFactory.__init__(self, url, debug=debug, debugCodePaths=debugCodePaths)
    self.clients = {}
    self.connector = DbConnector()
    self.messages = MessageParser()
    self.manager = LoginManager()
def register(self, client):
    print "%s connected" % client.peerstr
def unregister(self, client):
    if self.clients.has_key(client):
        self.processLogout(client)
    print "%s disconnected" % client.peerstr
def processMessage(self, client, msg):
    try:
        msg = self.messages.parseMessage(msg)
        action = msg['Type']
    except ValueError, e:
        logging.warning("[Parse]:%s", e.message)
        client.sendMessage(self.messages.createErrorMessage("could not parse your message"))
        return
    if action == "ChatMessage":
        self.processChatMessage(client, msg)
    # elif action == "Login":
    #   self.processLogin(client, msg)
    # elif action == "Logout":
    #   self.processLogout(client)
    elif action == "OpenId":
        self.manager.processLogin(client,msg)

def processChatMessage(self, client, msg):
    if not self.clients.has_key(client):
        client.sendMessage(self.messages.createErrorMessage('Not authorized'))
        return
    if not msg['Message']:
        client.sendMessage(self.messages.createErrorMessage('Invalid Message'))
        return
    if not msg['Recipient']:
        client.sendMessage(self.messages.createErrorMessage('Invalid Recipient'))
        return
    if msg['Recipient'] in self.clients.values():
        for c in self.clients:
            if self.clients[msg['Recipient']]:
                c.sendMessage(self.messages.chatMessage(msg['Sender'], msg['Message']))
                print "sent message from %s to %s: '%s' .." % (msg['Sender'], msg['Recipient'], msg['Message'])
    else:
        client.sendMessage(self.messages.createErrorMessage('User not registered'))
def checkSender(self, user, client):
    if user in self.clients.values() and self.clients[client] == user:
        return
    else:
        self.clients[client] = user

一个独立的html/js客户端可以连接和发送聊天消息。但是我想在打开websocket之前实现一个开放的id身份验证(由服务器执行)。

这是onload函数:

var wsuri = "ws://192.168.0.12:9000";
if ("WebSocket" in window) {
    sock = new WebSocket(wsuri);
} else if ("MozWebSocket" in window) {
    sock = new MozWebSocket(wsuri);
} else {
    log("Browser does not support WebSocket!");
    window.location = "http://autobahn.ws/unsupportedbrowser";
}
if (sock) {
    sock.onopen = function () {
        log("Connected to " + wsuri);
    }
    sock.onclose = function (e) {
        log("Connection closed (wasClean = " + e.wasClean + ", code = " + e.code + ", reason = '" + e.reason + "')");
        sock = null;
    }
    sock.onmessage = function (e) {
        receive(e.data);
    }
}

因为我是新的python/twisted我不知道如何做到这一点和例子大多只显示websocket聊天室没有身份验证。

如何正确实现openid ?因为它还需要重定向，这会破坏ws连接。

Answer 1

不能在重定向前打开ws。之后打开它，然后你的代码就可以工作了。

好运。

顺便说一下，谷歌在它的主页上也这样做。