Question

我试图将套接字对象提供给我的＆＃39; ConnectionHandler＆＃39; class，但是当使用这个套接字对象时，它会出现这样的错误：＆＃39;无法读取未定义的socket.io＆＃39;的属性套接字。

服务器类：

＆＃13;

Server.prototype.handleConnections = function ()
{
	this.queueTime = 15; // Queue time in seconds

	var that = this; // Create a global variable of the server object

	// On incoming connection
	this.io.on('connection', function (socket) {
		console.log('connection incoming...'); // Log a message to the server console

		// When a client tries to join the queue
		socket.on('client_join_queue', function (username) {
			// Check if the username is valid
			if (! (username.length < 3)) {
				var newPlayer = new player(username);
				var connectionHandler = new connectionHandling(socket, that, newPlayer);
				that.connections.push(connectionHandler);
			}
		});
	});
}

＆＃13;

ConnectionHandler类：

＆＃13;

'use strict';

var ConnectionHandler = function (_socket, _server, _player)
{
	this.socket = _socket;
	this.server = _server;
	this.player = _player;

	this.server.queueHandler.addPlayer(this.player);

	this.server.connections[0].socket.emit('player_joined_queue', this.player, this.server.queueHandler.getQueue().length);

	var that = this;
	socket.on('disconnect', function () {
		console.log("user disconnected");
		console.log("queue:", that.server.queueHandler.getQueue());
	});
}

module.exports.ConnectionHandler = ConnectionHandler;

＆＃13;

我完全不知道我做错了什么。

Answer 1

fljs在说什么，你发出一个名为'player_joined_queue'的事件。但是您正在侦听事件'client_join_queue'。您需要使用您正在发出的相同名称来侦听事件。所以你需要改变一个或另一个。例如，

socket.on('player_joined_queue', function (username) {
...

Javascript Socket.IO - 传递给其他对象时Socket为空

1 个答案: