Question

当X用户尝试与Y用户创建房间时。我这样做：

socket.join(room.roomId);
UserConnectionId.findOne({userId: data.opponentId}).then((connection) => {
    //connection.connectionId is user's socket Id
    var socketList = io.sockets.server.eio.clients;
    if (socketList[connection.connectionId] != undefined){                    
        socketList[connection.connectionId].join(room.roomId);
    }
});

但显示此错误：

UnhandledPromiseRejectionWarning：TypeError：socketList [connection.connectionId] .join不是函数

如何通过socketId获取套接字对象，以及如何在房间中加入此连接？

Answer 1

首先，您需要正确处理错误，我不知道 socketList [connection.connectionId] .join（room.roomId）的作用。尝试使用此代码段可正确处理错误。然后使用 socketList [connection.connectionId] .join（room.roomId）函数的手册进行检查

socket.join(room.roomId);
UserConnectionId.findOne({userId: data.opponentId})
.then((connection) => {
    //connection.connectionId is user's socket Id
    var socketList = io.sockets.server.eio.clients;
    if (socketList[connection.connectionId] != undefined){                    
        return socketList[connection.connectionId].join(room.roomId);
    }
});
.catch((errors) => {
    console.log(errors)
 })

Answer 2

解决方案*

我变了 var socketList = io.sockets.server.eio.clients; 与 var socketList = io.sockets.sockets; 它是固定的

如何在房间里加入对手

2 个答案: