3

我正在使用 node.js 来实现 websocket 服务器和客户端。他们之间的握手是这样的。

请求网址：ws://localhost:8015/

请求方法：GET

状态码：101 交换协议

请求标头

Cache-Control: no-cache
Connection: Upgrade
Cookie: SQLiteManager_currentLangue=2
Host: localhost:8015
Origin: http:/localhost:8080
Pragma: no-cache
Sec-WebSocket-Extensions: x-webkit-deflate-frame
Sec-WebSocket-Key: A/knWtXFtTa5V6po8XOfjg==
Sec-WebSocket-Protocol: echo-protocol
Sec-WebSocket-Version: 13
Upgrade: websocket
User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_8_3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/27.0.1453.116 Safari/537.36

响应标头

Connection: Upgrade
Origin: http:/localhost:8080
Sec-WebSocket-Accept: 5OUv+g5mBPxVDug4etJfGX4lxIo=
Sec-WebSocket-Protocol: echo-protocol
Upgrade: websocket

服务器正在获取从客户端发送的消息（我正在控制台上记录消息），但是当服务器发送消息时，客户端中的 onmessage 事件不会被触发。让我感到困惑的另一件事是，一旦打开连接，客户端中的 onmessage 事件只会触发一次。

请帮助...我正在尝试将服务器上的消息回显给客户端。

编辑：

这就是我在 websocket 客户端中处理事件的方式......

html5WebSocketClient.connect = function(){
     if(window.WebSocket != undefined){
         if(connection.readyState == undefined || connection.readyState > 1)
             connection = new WebSocket('ws://localhost:8015/','echo-protocol');
     }
     if (window.MozWebSocket) {
          window.WebSocket = window.MozWebSocket;
     }

 connection.onopen = html5WebSocketClient.onOpen(event);
 connection.onmessage = html5WebSocketClient.onMessage(event);
 connection.onclose = html5WebSocketClient.onClose(event); 
 connection.onerror = html5WebSocketClient.onError(event);

};

html5WebSocketClient.onOpen = function(event)
{
    $('#some_label').text("Connected");
};

html5WebSocketClient.onClose = function(event)
{
    $('#some_label').text("Disconnected");
};

html5WebSocketClient.onError = function(event)
{
    $('#some_label').text("Error");
};

//This is only getting fired when connection opens
html5WebSocketClient.onMessage = function(message)
{
    if(message.data != undefined)
        {
        alert($.parseJSON(message.data));
    }
};

//Server is getting this message
html5WebSocketClient.sendMessage = function()
{
   var message = {"name": value, "name": value};
   connection.send(JSON.stringify(message));
};

这就是我实现服务器的方式..

var http = require('http');
var WebSocketServer = require('websocket').server;

var server = http.createServer(function (req, res) {
  res.writeHead(200, {'Content-Type': 'text/plain'});
  res.end('Hello Node.js From HTML5\n');
}).listen(8015, "127.0.0.1");

wsServer = new WebSocketServer({
    httpServer: server,
    autoAcceptConnections: false
});

wsServer.on('request', function(request) {

    var connection = request.accept('echo-protocol', request.origin);
    console.log((new Date()) + ' Connection accepted.');
    connection.on('message', function(message) {
        if (message.type === 'utf8') {
            console.log('Received Message: ' + message.utf8Data);
            connection.send(message.utf8Data); //Client is not getting this message..?
        }
    });
    connection.on('close', function(reasonCode, description) {
        console.log((new Date()) + ' Peer ' + connection.remoteAddress + ' disconnected.');
    });
});
4

1 回答 1

6

这些行负责您的问题：

connection.onopen = html5WebSocketClient.onOpen(event);
connection.onmessage = html5WebSocketClient.onMessage(event);
connection.onclose = html5WebSocketClient.onClose(event); 
connection.onerror = html5WebSocketClient.onError(event);

让我们来分析一下。使用参数（即）html5WebSocketClient.onOpen(event);调用并返回（不返回任何内容）。从而变成。但应该是一个功能。因此，不要调用这些函数，只需执行以下操作：onOpeneventundefinedundefinedonOpenconnection.onopenunedefinedconnection.onopen

connection.onopen = html5WebSocketClient.onOpen;
connection.onmessage = html5WebSocketClient.onMessage;
connection.onclose = html5WebSocketClient.onClose; 
connection.onerror = html5WebSocketClient.onError;
于 2013-07-13T21:57:37.940 回答