从单一实例处理程序类更改 UI

Question

>我有以下单例处理程序类

public class MyHandler
{
private static Handler handler;
private static boolean isRunning;
public static Handler getHandler(Runnable myRunnable)
{
    if (handler == null)
    {
        initHandler(myRunnable);
    }
    return handler;
}

private static void initHandler(Runnable myRunnable)
{
    handler = new Handler();
    isRunning = true;
    handler.postDelayed(myRunnable, 5000);
}
public static void reRunHandler(Runnable myRunnable)
{
    isRunning = true;
    handler.postDelayed(myRunnable, 45000);
}
public static void stopMyHandler()
{
    isRunning = false;
    handler.removeCallbacksAndMessages(null);
}
}

但是，如何从此处更新我的 UI？因为可运行对象在我的活动中。显然我无法使用getHandleMessage与它进行通信。

如果你需要更多的代码，我如何使用这个，我可以分享。

Answer 1

这很简单：

new Handler(Looper.getMainLooper()).post(new Runnable() {           
    @Override
    public void run() {
        //do whatever you want on the UI thread
    }
});

Answer 2

句柄具有用于此目的的函数：

private final Handler handler =  new Handler() {
    public void handleMessage(Message msg) {
        // here you can get data from Message and update your UI. runs in UI thread
    }
};

如果要将包含数据的消息发送到处理程序，请使用下一个代码：

Message m = new Message();
Bundle b = new Bundle();
b.putInt("myNumber", 5); // for example
m.setData(b);
myHandler.sendMessage(m);