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XMPP aSmack - 如何获取当前用户状态(离线/在线/离开/等)？

我正在开发基于asmack lib的Android上的聊天应用程序.我在ListView上显示所有用户,但我使用图像显示在线/离线用户.但它只返回离线图像,即使用户在线,这里是我的代码

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    // setContentView(R.layout.buddies);

    Controller.getInstance().roster = Controller.getInstance().connection.getRoster();
//  ArrayList<Buddy> buddies = new ArrayList<Buddy>();
    Collection<RosterEntry> entries = Controller.getInstance().roster.getEntries();
    Controller.getInstance().buddyList = new Buddy[entries.size()];

    int i = 0;

    for (RosterEntry r : entries) {
        Buddy bud = new Buddy();
        VCard card = new VCard();
        try {
            ProviderManager.getInstance().addIQProvider("vCard",
                    "vcard-temp", new VCardProvider());
            card.load(Controller.getInstance().connection, r.getUser());
        } catch (XMPPException e) {
            Log.e("ChatOnAndroid", e.getMessage() + " " + r.getUser() + " "
                    + e.getLocalizedMessage());
        }
        bud.jid …