Question

环境

socket.io 2.1.1 
express 4.16.3  
Node v10.13.0

期望的行为

当客户加入房间时，房间中的所有客户都会收到房间中所有客户的列表，包括每个客户的自定义socket.username属性。

我尝试过的事情

尝试01：

将clients()方法与回调一起使用，请参见：

https://socket.io/docs/server-api/#namespace-clients-callback。

socket.on('join', function(data) {
    // get desired room name
    var room = data.room;
    // get username
    var username = data.username;
    // add username key:value to socket
    socket.username = username;
    // add socket to room
    socket.join(room);

    // get list of clients in room
    io.of('/').in(room).clients((error, clients) => {
        if (error) {
            throw error;
        } else {
            // send data to all clients in room
            io.in(room).emit('room_join', { username: socket.username, room: room, clients: clients });
        }
    });
});

在上面的示例中，clients作为套接字ID的数组返回。

尝试02：

使用受this answer启发的io.sockets.adapter.rooms[room]：

socket.on('join', function(data) {
    var room = data.room;
    var username = data.username;
    socket.username = username;
    socket.join(room);
    var sockets_in_room = io.sockets.adapter.rooms[room];
    io.in(room).emit('room_join', { username: socket.username, room: room, sockets_in_room: sockets_in_room });
});

以上返回：

Room { sockets: { '3-fHYqg7e8h-6K8EAAAC': true }, length: 1 }

问题

似乎可以通过以下方式将自定义username属性添加到套接字：

socket.username= "john";

如何在“客户端对象”列表中访问该属性，以便可以访问房间中所有客户端的用户名？

编辑1

在相关的情况下，在这种情况下可能有数千个房间，每个房间名称都来自唯一的内容ID，因此任何解决方案都需要适应该约束。

编辑2

我进行了以下操作，但似乎可行，不确定是否朝正确的方向前进

var users_in_rooms = {};

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

    // handle joining
    socket.on('join', function(data) {
        var room = data.room;
        var username = data.username;
        socket.username = username;
        socket.join(room);
        // add key:value to users_in_rooms object
        users_in_rooms[socket.id] = username; 

        // get list of clients in room
        io.of('/').in(room).clients((error, clients) => {
            if (error) {
                throw error;
            } else {
                // for each socket id in the clients array, look up its corresponding username
                var client_object_array = clients.map(socketid => ({
                    socketid: socketid,
                    username: users_in_rooms[socketid]
                }));
                // return client_object_array to all clients
                io.in(room).emit('room_join', { client_object_array: client_object_array, username: socket.username, room: room  });
            }
        });
    });

    // handle leaving
    socket.on('leave_all', function() {
        socket.leaveAll();
        // when a user leaves all rooms, delete their corresponding entry from the users_in_rooms object
        delete users_in_rooms[socket.id];
    });

    // handle disconnecting
    socket.on('disconnect', function(reason) {
        console.log("client " + socket.id + " disconnected because:");
        console.log(reason);
        delete users_in_rooms[socket.id];
    });

});

如何在socket.io中将用户名属性添加到客户端对象？

0 个答案: