socket.io: only one room to a socket

Question

I would like to guarantee that a user is only in one room at a time, but I am having trouble implimenting this. I have boiled the issue down to not being able to get a list of all the rooms a socket is currently in.

My current strategy is to do something like this:

io.on('connection', function(socket) {
    console.log('A user just connected with id ' + socket.id);
    socket.on("change room", function(data) {
        // this SHOULD return a list of all rooms this user is in
        var socketsRooms = socket.rooms;
        console.log(socket.rooms) //{ '/#3tE4Up5PRbTfsU0JAAAD': '/#3tE4Up5PRbTfsU0JAAAD' }
        console.log(typeof socket.rooms) // object
        // Loop through the rooms and leave them
        for (var room in socketsRooms) {
            console.log('LEAVING ROOM: ', room);
            socket.leave(room);
        }

        // Join the new room
        socket.join(data.newroom);
    });
    socket.on('disconnect', function() {
        console.log('user ' + socket.id + ' disconnected');
    });
})

Is there an issue in the docs with socket.rooms? It says it should return an array of rooms that a socket is connected to, but I get back data as an id object. EX: { '/#3tE4Up5PRbTfsU0JAAAD': '/#3tE4Up5PRbTfsU0JAAAD' }

... In my case I would expect something like ['issues']

On the Socket.io github page the socket.rooms description stated:

A hash of strings identifying the rooms this socket is in, indexed by room name.

BEFORE MARKING AS A DUPLICATE:

I have already checked out the folowing questions:

1. How to leave all rooms that the socket is connected to at one go in Node.js
2. socket.io: assure socket joins only one room at a time

The answer to the first question seems the most hacked together solution the I am pretty sure it will work because it assumes the solution as maintaining our own list, but I would prefer a working solution using socket.io over that method.

I am using the latest verion of socket.io v1.4.5 with angular2 beta on the client (if that helps out at all)

Answer 1

In socket.io, each socket is also connected to its own room. This room has the same name of socket.id . So when leaving the rooms, make sure you don't leave its own room.

for(room in socket.rooms){
    if(socket.id !== room) socket.leave(room);
}
socket.join(data.newroom, function(){
  console.log('rooms', socket.rooms); // here you'll see two rooms: one with socket.id and another with data.newroom
});