Android

Question

我正在为我的聊天应用程序使用 aSmack 和 Openfire。

我已经编写了监听互联网连接并连接到 XMPP 服务器的后台服务。

当服务尝试连接 XMPP 服务器时，它抛出 'Already Logged in to server' 异常。

我在服务器设置中将资源策略设置为 'Always kicked' 并尝试将 'null' 作为 login() 方法中的第三个参数传递，但其中 none 对我有用.谁能帮帮我？

04-11 18:09:50.293: E/AndroidRuntime(25422): FATAL EXCEPTION: Thread-2212
04-11 18:09:50.293: E/AndroidRuntime(25422): Process: com.example.singlechat:remote, PID: 25422
04-11 18:09:50.293: E/AndroidRuntime(25422): java.lang.IllegalStateException: Already logged in to server.
04-11 18:09:50.293: E/AndroidRuntime(25422):    at org.jivesoftware.smack.XMPPConnection.login(XMPPConnection.java:232)
04-11 18:09:50.293: E/AndroidRuntime(25422):    at org.jivesoftware.smack.Connection.login(Connection.java:371)
04-11 18:09:50.293: E/AndroidRuntime(25422):    at com.example.singlechat.BackgroundService.run(BackgroundService.java:173)
04-11 18:09:50.293: E/AndroidRuntime(25422):    at java.lang.Thread.run(Thread.java:818)

连接代码：

// Connect XMPP Server
public void connectXMPPServer() {

    Thread t = new Thread(new Runnable() {

        @Override
        public void run() {

            // Create a connection
            ConnectionConfiguration connConfig = new ConnectionConfiguration(
                    HOST, PORT, SERVICE);
            connConfig.setReconnectionAllowed(true);

            xmppConnection = new XMPPConnection(connConfig);

            try {

                xmppConnection.connect();
                Log.i("XMPPChatDemoActivity", "Connected to "
                        + xmppConnection.getHost());
            } catch (XMPPException ex) {

                Log.e("XMPPChatDemoActivity", "Failed to connect to "
                        + xmppConnection.getHost());
                Log.e("XMPPChatDemoActivity", ex.toString());
            }

            try {

                if (xmppConnection.isConnected()) {

                    xmppConnection.login(USERNAME, PASSWORD);
                    Log.i("XMPPChatDemoActivity", "Logged in as "
                            + xmppConnection.getUser());

                    Presence presence = new Presence(
                            Presence.Type.available);
                    xmppConnection.sendPacket(presence);
                }
            } catch (XMPPException ex) {

                Log.e("XMPPChatDemoActivity", "Failed to log in as "
                        + USERNAME);
                Log.e("XMPPChatDemoActivity", ex.toString());
            }
        }
    });
    t.start();
}

// Broadcast receiver; listens to network state
public BroadcastReceiver networkStateReceiver = new BroadcastReceiver() {

    @Override
    public void onReceive(Context context, Intent intent) {

        try {

            if (isNetworkOn()) {

                Log.i("BackgroundService-BroadcastReceiver",
                        "Network is on");
                connectXMPPServer();
            } else if (!isNetworkOn()) {

                Log.i("BackgroundService-BroadcastReceiver",
                        "Network is off");
                xmppConnection.disconnect();
            }
        } catch (Exception e) {

            Log.e("BackgroundService-networkStateReceiver", e.toString());
        }
    }
};

// Returns network state
public boolean isNetworkOn() {

    ConnectivityManager connMngr = (ConnectivityManager) getApplicationContext()
            .getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo netInfo = connMngr.getActiveNetworkInfo();
    return (netInfo != null && netInfo.isConnected());
}

Answer 1

Openfire 提供了很多自定义选项，例如 Kick resources after logged out 等。但是 none 对我有用。所以 try-catch 是剩下的选项。如果有人做更好的研究，他们的建议将不胜感激。

try {

    xmppConnection.login(USERNAME, PASSWORD);
} catch (IllegalStateException e) {

    Log.e("MessagingService", "Already Logged in as " + xmppConnection.getUser());
}
Log.i("MessagingService", "Logged in as " + xmppConnection.getUser());

Answer 2

这是 Smack 库的问题 - XMPP Presence 节在启用扩展时始终发送，无论设置如何。

查看开发者主题：https://community.igniterealtime.org/thread/55778

您可以通过从 app/build.gradle 中删除 Smack 扩展依赖来验证这一点。如果您不再有异常，则意味着 Smack 开发人员应该提供解决方案。

Android - XMPP：'Already logged in to server' 异常

Android - XMPP : 'Already logged in to server' exception

exception

login

xmpp