因此,每次刷新页面时,似乎sockjs正在创建一个新连接。
我在每个channel.onmessage上将每条消息保存到我的mongodb,所以如果我刷新我的页面7次并发送消息,我会将7条相同内容的消息保存到我的mongodb中。
这是非常有问题的,因为当我进入聊天室时检索这些消息时,为了查看日志,我会看到一堆重复的消息。
我想跟踪所有“活动”的连接,如果用户尝试建立另一个连接,我希望能够从服务器端强制关闭旧连接,因此只有1个连接一次收听每条消息
我该怎么做?
var connections = {};
//creating the sockjs server
var chat = sockjs.createServer();
//installing handlers for sockjs server instance, with the same url as client
chat.installHandlers(server, {prefix:'/chat/private'});
var multiplexer = new multiplexServer.MultiplexServer(chat);
var configChannel = function (channelId, userId, userName){
var channel = multiplexer.registerChannel(channelId);
channel.on('connection', function (conn) {
// console.log('connection');
console.log(connections);
connections[channelId] = connections[channelId] || {};
if (connections[channelId][userId]) {
//want to close the extra connection
} else {
connections[channelId][userId] = conn;
}
// }
// if (channels[channelId][userId]) {
// conn = channels[channelId][userId];
// } else {
// channels[channelId][userId] = conn;
// }
// console.log('accessing channel! ', channels[channelId]);
conn.on('new user', function (data, message) {
console.log('new user! ', data, message);
});
// var number = connections.length;
conn.on('data', function(message) {
var messageObj = JSON.parse(message);
handler.saveMessage(messageObj.channelId, messageObj.user, messageObj.message);
console.log('received the message, ', messageObj.message);
conn.write(JSON.stringify({channelId: messageObj.channelId, user: messageObj.user, message: messageObj.message }));
});
conn.on('close', function() {
conn.write(userName + ' has disconnected');
});
});
return channel;
};
答案 0 :(得分:0)
只需使用.close
:
if (connections[channelId][userId]) {
// want to close the extra connection
connections[channelId][userId].close();
} else {
connections[channelId][userId] = conn;
}