Cant join a room

+1 vote

Im using Android. I have two Android devices. The first is creating a room. The second tries to

join that room by invoking "joinRoomInRange". But surprise surprise: It doesnt work somehow.

I checked if theres any room available by invoking "getAllRooms". I saw that the count of rooms is zero.

In the function "onJoinRoomDone" the event code will always be 2, so it will always create a new room.

I would really appreciate your help!


Heres some code:


The Client class:

public class Client

    public static WarpClient myGame;

    public static String userID = "0";

    public static Runnable runConnection = new Runnable() {
        public void run() {

    public static void connect()

        try {
            myGame = WarpClient.getInstance();
            myGame.addConnectionRequestListener(new MyConnectionListener());
            myGame.addChatRequestListener(new MyChatListener());
            myGame.addZoneRequestListener(new MyZoneListener());
            myGame.addRoomRequestListener(new MyRoomRequestListener());
            myGame.addNotificationListener(new MyNotificationListener());

            new Thread(runConnection).start();
        catch(Exception e)




Function of ConnectionRequestListener class that implements ConnectionRequestListener:

    public void onConnectDone(ConnectEvent event) {

        if(event.getResult() == WarpResponseResultCode.SUCCESS){
            System.out.println("yipee I have connected");






Function of RoomRequestListener class that implements RoomRequestListener:


    public void onJoinRoomDone(RoomEvent event) {

        if (event.getResult() == WarpResponseResultCode.SUCCESS) {
            // success case
            roomId = event.getData().getId();

        } else if (event.getResult() == WarpResponseResultCode.RESOURCE_NOT_FOUND) {
            // no such room found
            //Create new one
            HashMap<String, Object> data = new HashMap<String, Object>();
            data.put("result", "");
            Client.myGame.createRoom("superjumper", "shephertz", 2,data);

        } else {



asked Mar 27, 2020 in Work In Progress by Sulp Sulpc (15 points)

1 Answer

0 votes

The solution was that the client that is creating the room also needs to join and subscribe it!

Otherwise the room is not available to other clients!

answered Mar 28, 2020 by Sulp Sulpc (15 points)
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