How to keep user joined to their multiple rooms socket.io - reactjs

Question

I am completely new to socket.io and I have been stuck on this problem for so many days. I can't find any deep youtube vidoes on this topic either. What I am trying to do is, I am creating a chat app. As soon as you log in you will see all the groups and you can join these groups. Let's assume that user has joined 3 groups. As soon as that user logs in I would like to join that user to those three groups so that he would get messages when there is any. How can that be done??
This is what I have done so far,
Front-end
when message is sent
socket.emit("message", { name, message, room: selectedGroup.name });
socket.emit("join-group", { name, room: group.name });
socket.on("message", ({ name, message, room }) => {
const update = { ...group };
update.messages.push({ sender: name, message });
setSelectedGroup(update);
});
Back-End
io.on("connection", (socket) => {
socket.on("join-group", ({ name, room }) => {
socket.join(room);
});
socket.on("message", ({ name, message, room }) => {
socket.join(room);
io.to(room).emit("message", { name, message, room });
});
});

Answer 1

my main language is not English, so I apologize for that
There is an adapter in the io object that can access all active rooms inside, and if you want to have access to all of them when the user logs in, you must use these commands in onConnection
However, versions 2,3,4 are different in this regard.
https://socket.io/docs/v4/adapter/
From the top right you can find the version you want
const socketio = require("socket.io");
const server = http.createServer(app);
const io = socketio(server);
const activeRooms(io)=> {
const arr = Array.from(io.sockets.adapter.rooms);
// [['room1', Set(2)], ['room2', Set(2)]]
const filtered = arr.filter(room => !room[1].has(room[0]))
// ['room1', 'room2']
const res = filtered.map(i => i[0]);
return res;
}
const rooms = activeRooms()
if (rooms) {
rooms.forEach((element) => {
socket.join(element.toString());
});
socket.join(socket.username);
}
}