Question

我想发送消息到特定房间（房间里的所有客户），我已经搜索了很多并尝试了这些选项：

        io.in(socket.RoomId).emit('Test');
        io.to(socket.RoomId).emit('Test');        
        io.sockets.in(socket.RoomId).emit('Test');

我想发送服务器，所以不使用房间里现有的套接字。但是因为我找不到办法，所以我尝试了以下方法：

this.ListOfSockets[0].to(RoomId).emit('Test');

这不起作用，所以我试着像这样向插座本身发射：

this.ListOfSockets[0].emit('Test');

这就像我预期的那样工作。

我做错了什么？为什么我不能作为服务器发送到特定房间的所有客户端？

更新2： 为li X添加更多代码

io.on('connection', function(socket){

console.log('client connected');
playerCount++;

//For testing purposes I give player random generated string using: var shortid = require('shortid');
socket.id = shortid.generate();
console.log('id of this socket is:' + socket.id);

//When client decides to play the game and want's to join a random room
socket.on('JoinRoom', function () 
{
    //Checks if there exist an open room
    if(OpenRoom.length > 0)
    {
        //Joins first room in openroom list
        socket.join(OpenRoom[0].RID);
        socket.RoomId = OpenRoom[0].RID;

        //Joinedroom lets player open the next scene in my game
        socket.emit('JoinedRoom');

        //Joinedroom is a function that sends information to the socket about the room and the state of the game,
        //adding the the player to room etc.
        JoinedRoom(socket.id, OpenRoom[0].RID, socket);
    }
    else
    {
        //If there is no open room, it creates a new room with new id.
        OpenRoom.push({ RID:shortid.generate(), TimeOutID: null });
        //Joins room
        socket.join(OpenRoom[OpenRoom.length - 1].RID);
        socket.RoomId = OpenRoom[OpenRoom.length - 1].RID;

        //Creates room and fills information to it
        socket.emit('JoinedRoom');
        var NewRoom = new Room(OpenRoom[OpenRoom.length - 1].RID)
        NewRoom.addPlayer(socket.id);
        NewRoom.addSocket(socket);

        Rooms.push(NewRoom);

        OpenRoom[OpenRoom.length - 1].TimeOutID = setTimeout(CloseRoom, 5000, OpenRoom[OpenRoom.length - 1]);
    }

    //Checking if RoomId exists, it exists.
    console.log(socket.RoomId);

    //When I send 'test' to client, the client should print 'Succesfully Tested' to the console

    //For io.to I dont get the message 'Succesfully Tested'
    io.to(socket.RoomId).emit('Test');

    //With socket.emit I get 'Succesfully Tested' message inside my console at the client side.
    socket.emit('Test');

});

提前致谢

Answer 1

默认情况下，在派生socket.id时，所有套接字都会连接到自己的个人房间，但要创建单独的房间，您需要在要成为房间一部分的套接字上调用data。

之后，您将能够使用socket.join('foo');向房间发送消息。另请注意，您还可以在io.to('foo').emit('test');和broadcast之间添加io.您可以向所有套接字发送消息，但是发出呼叫的套接字。

.to 这也应该是 socket.join(OpenRoom[OpenRoom.length - 1].RID);

您需要使用socket.join(OpenRoom[OpenRoom[0]].RID);在此范围内，您可以调用onTest函数并通过数据传递必要的信息。

无法向特定房间发送消息[socket.io]

1 个答案: