有很多方法可以做到这一点，但最好的是使用如下处理程序

long millisecDelay=3000

Handler().postDelayed({
  // do your work here
 },millisecDelay)

final Handler handler = new Handler(); 
Timer t = new Timer(); 
t.schedule(new TimerTask() { 
    public void run() { 
        handler.post(new Runnable() { 
            public void run() { 
                //DO SOME ACTIONS HERE , THIS ACTIONS WILL WILL EXECUTE AFTER 5 SECONDS...
            }
        }); 
    } 
}, 5000); 

谢谢你的回答，我找到了一个最适合我需要的解决方案。

Handler myHandler = new DoSomething();
Message m = new Message();
m.obj = c;//passing a parameter here
myHandler.sendMessageDelayed(m, 1000);

class DoSomething extends Handler {
    @Override
    public void handleMessage(Message msg) {
      MyObject o = (MyObject) msg.obj;
      //do something here
    }
}

你可以使用这个最简单的解决方案:

new Handler().postDelayed(new Runnable() {
    @Override
    public void run() {
        //Write your code here
    }
}, 5000); //Timer is in ms here.

否则，下面可能是另一个干净有用的解决方案:

new Handler().postDelayed(() -> 
{/*Do something here*/}, 
5000); //time in ms

我更喜欢使用View.postDelayed()方法，下面是简单的代码:

mView.postDelayed(new Runnable() {
    @Override
    public void run() {
        // Do something after 1000 ms
    }
}, 1000);

对于简单行句柄Post延迟，您可以执行以下操作:

new Handler().postDelayed(new Runnable() {
    @Override
    public void run() {
        // Do someting
    }
}, 3000);

我希望这对你们有帮助