创建NoResponseException时获取MultiUserChat

Question

我正在尝试用XMPP创建一个多用户聊天(4.1.4)。我建立了一个登录连接

XMPPTCPConnectionConfiguration.Builder configBuilder = XMPPTCPConnectionConfiguration.builder();
configBuilder.setSecurityMode(ConnectionConfiguration.SecurityMode.disabled);
    configBuilder.setResource(RESOURCE);
    configBuilder.setServiceName(DOMAIN);
    configBuilder.setHost(HOST);
    XMPPTCPConnection connection = new XMPPTCPConnection(configBuilder.build());
    connection.addConnectionListener(connectionListener);

AsyncTask<Void, Void, Boolean> connectionThread = new AsyncTask<Void, Void, Boolean>() {
        @Override
        protected Boolean doInBackground(Void... arg0) {
            try {
                connection.connect();
                if(connection != null) {
                PingManager mPingManager = PingManager.getInstanceFor(connection);
                mPingManager.setDefaultPingInterval(1000 * 50);
                mPingManager.setPingInterval(1000 * 50);
                connection.setPacketReplyTimeout(1000 * 50);
                mPingManager.registerPingFailedListener(new PingFailedListener() {
                @Override
                public void pingFailed() {
                    // TODO Auto-generated method stub
                    Log.e("PING", "ping failed");
                    // AGain Reconnecting code here..
            }
        });
    }
            } catch (IOException | SmackException | XMPPException e) {
                Log.d(LOCAL_TAG, "connectConnection  exc: "+e.getMessage());
            }
            return null;
        }
    };
    connectionThread.execute();

而且它成功地连接了起来。连接之后，我试着用，

    try {
        SharedPreferences mPreferences = context.getSharedPreferences("App_Pref", Context.MODE_PRIVATE);
        String userJid = userJidAddress(with local ip);
        String userName =  "Test5";
        Log.d(LOCAL_TAG, "createGroupChat  jid: "+userJid+" -- connection.getServiceName(): "+connection.getServiceName());

        MultiUserChatManager manager = MultiUserChatManager.getInstanceFor(connection);
        MultiUserChat muc = manager.getMultiUserChat(userJid);

        muc.create("InstantRoom");

        Log.d(LOCAL_TAG, "createGroupChat  -- Group CEATED Successfully ");
        Form form = muc.getConfigurationForm();
        Form submitForm = form.createAnswerForm();

        List<FormField> fields = form.getFields();
        Log.d(LOCAL_TAG, "createGroupChat  -- fields.size(): "+fields.size());
        for (int i = 0; i < fields.size(); i++) {
            FormField field = (FormField) fields.get(i);
            if (!FormField.Type.hidden.equals(field.getType()) && field.getVariable() != null) {
                submitForm.setDefaultAnswer(field.getVariable());
            }
        }

        List owners = new ArrayList();
        owners.add("test5"+connection.getServiceName());
        owners.add("test7"); //Another user
        submitForm.setAnswer("muc#roomconfig_roomowners", owners);
        submitForm.setAnswer("muc#roomconfig_persistentroom", true);
        muc.sendConfigurationForm(submitForm);
        muc.join("InstantRoom");

    }
    catch(Exception e){
        Log.d(LOCAL_TAG, "createGroupChat --  Exception: "+e.toString());
    }

但是在创建这个组时，我得到了一个异常

    "org.jivesoftware.smack.SmackException$NoResponseException: No response received within reply timeout. Timeout was 50000ms (~50s). Used filter: AndFilter: (FromMatchesFilter (full): test5@172.21.4.199/instantroom, StanzaTypeFilter: org.jivesoftware.smack.packet.Presence)." 

如果我增加PingIntervals和PacketReplyTimeout毫秒，那么我会收到如下消息

"Thread5，tid=4403，WaitingInMainSignalCatcherLoop，Thread*=0xb8e933a8，peer=0x12c000a0，信号守护者“：对信号3的反应。。

没有异常和应用程序设置强制自动关闭。希望如此，我在muc.create("InstantRoom")行中得到了异常，因为在该行之后我无法获得日志。有人能帮我找出我做错了什么，为什么我会犯这个错误，以及如何在此之后继续下去。提前谢谢。

Answer 1

这两行错了

 MultiUserChat muc = manager.getMultiUserChat(userJid);

        muc.create("InstantRoom");

当您得到一个MUC时，您必须得到一个名称(这样经理就可以告诉您是否已经存在)。

实际上，当您创建一个muc时，您必须输入一个用户名。

所以

String myMUCName = "InstantRoom";
String myMUCService = "conference.myServer";
String myMUCfullName = myMUCName + "@" + myMUCService;
MultiUserChat muc = manager.getMultiUserChat(myMUCfullName);
muc.create(userName);

(如果这不能解决您的问题，请在向您保证它已经在服务器上注册的服务之后告诉我--也许您在表单上会有额外的问题)