NodeJS - socket.broadcast.to不在嵌套侦听器内部发射

Question

我试图向特定的房间发出消息，一旦＃34;加入房间＆＃39;监听器由客户端触发，如果我将代码放在joinedRoom监听器之外，代码工作正常，否则它什么都不做。

代码：

app.get('/room/:room/user/:user', function(req, res){
    var room = {
        username: req.params.user,
        roomname: req.params.room
    };
    res.render('room', room);
});

var users  = {};

io.sockets.on('connection', function (socket) {
    socket.on('joinedRoom', function(roomData){
        socket.username = roomData.username;
        socket.room = roomData.roomname;

        console.log("Roomname: " + socket.room);
        console.log("Username: " + socket.username);

        socket.join(socket.room);
        socket.broadcast.to(socket.room).emit('newUser', socket.username);

        socket.on('disconnect', function(){
            socket.broadcast.emit('userLeft', socket.username);
            socket.leave(socket.room);

            console.log('Connection id: ' + socket.id);
        });
    });
});

我看到了文档和一些示例代码，所有内容都是正确的（当它只是语法时）我在这里错过了一些简单的东西吗？

谢谢！

修改

客户代码：

var socket, roomname, ioRoom;
var socket = io.connect('http://localhost:3000');

socket.on('enterRoom', function(roomname){
    console.log("ENTERED ROOM: " + roomname);
});

socket.on('newUser', function(username){
    pushUserName(username);
    pushUserStatus(username, ' has joined the room <br/>')
});

socket.on('newRoom', function(data){
    alert(data)
});

socket.on('userLeft', function(username){
    pushUserStatus(username, ' has left the room <br/>')
})

function pushUserName(username){
    var el = document.getElementById("username");
    el.innerHTML += username + '<br/>';
}

function pushUserStatus(username, message){
    var el = document.getElementById("joined");
    el.innerHTML += username + message;
}