Java/android 如何在延迟 3 秒后启动 AsyncTask？

Question

如何在延迟 3 秒后启动 AsyncTask？

Answer 1

Using handlers as suggested in the other answers, the actual code is:

new Handler().postDelayed(new Runnable() {
    @Override
    public void run() {
        new MyAsyncTask().execute();
    }
}, 3000);

Answer 2

You can use Handler for that. Use postDelayed(Runnable, long) for that.

Handler#postDelayed(Runnable, Long)

Answer 3

You can use this piece of code to run after a 3 sec delay.

new Timer().schedule(new TimerTask() {          
    @Override
    public void run() {

        // run AsyncTask here.    


    }
}, 3000);

Answer 4

使用 Handler 类，并定义 Runnable handleMyAsyncTask将包含在 3000 毫秒延迟后执行的代码：

mHandler.postDelayed(handleMyAsyncTask, 1000*3);

Answer 5

Use CountDownTimer.

  new CountDownTimer(3000, 1000) {

        public void onTick(long millisUntilFinished) {

           //do task which continuously updates

        }

        public void onFinish() {

           //Do your task
         
        }

    }.start();

3000 is total seconds and 1000 is timer tick on that time means on above case timer ticks 3 time.