创建 Xmpp 连接并在不重新创建的情况下传递实例的最佳方法

Question

我有 class 称为 XmppChatManager，它具有所有必要的方法，如 initializingConnection() 和 sendOneToOneMessage()，其他 class 需要使用此 class 需要做这样的事情：

XmppChatManager xmppchatmanager = new XmppChatManager(this, userId, password);

runOnUiThread(new Runnable() {
                public void run() {
                    Thread thread = new Thread(){
                        public void run(){
                            try {
                                xmppchatmanager.initializingConnection();
                            } catch (IOException e) {
                                e.printStackTrace();
                            } catch (XMPPException e) {
                                e.printStackTrace();
                            } catch (SmackException e) {
                                e.printStackTrace();
                            }
                        }
                    };
                    thread.start();
                }
            });

This is pretty much all I need to get started, the issue i'm having is when I try calling the sendOneToOneMessage() method from the chatClass, I have to get the instance of the created connection because Smack Error will occur if I try reconnecting when the connection is already initiated! So I need to make available the instance of that initiated connection by creating a method called getInstance() I guess this might be simple but I have tried but I'm not getting it. Any one with better way of handling such?

Answer 1

让你的 xmppchatmanager 成为单例，例如

public class XmppChatManager{
   static class Loader{
        static XmppChatManager sInstance = new XmppChatManaget();
   }

   public static XmppChatManager getInstance(){
        return Loader.sInstance;
   }
}