Question

我正在编写一个Android应用程序，它允许有人添加两个数字并输入答案。但我希望这些数字只显示5秒，然后显示一个新数字，如果他们输入正确或错误的答案，计时器重置并显示新数字.. 我已经编写了执行随机数的代码，而其他只是计时器无法执行有人请帮忙

Answer 1

使用Handler和Runnable应该适合您，但不要使用Anonymous runnable，因为它们可能导致内存泄漏。而是将runnable扩展为静态类，并在removeCallbacks中使用onDestroy。

此外，您可以使用WeakReference，因为onDestroy不能保证被调用，因此如果您的活动被杀，WeakReference将允许GC释放内存

public class BarActivity extends AppCompatActivity {

    private Handler mHandler;
    private FooRunnable mRunnable;

    private void finishActivityAfterDelay(int milliSeconds) {
        mHandler = new Handler();
        mRunnable = new FooRunnable(this);
        mHandler.postDelayed(mRunnable, 5000); // 5 seconds
    }

    @Override
    protected void onDestroy() {
        mHandler.removeCallbacks(mRunnable);
        super.onDestroy();
    }

    private static class FooRunnable implements Runnable {
        private WeakReference<AppCompatActivity> mWeakActivity;

        public FooRunnable(AppCompatActivity activity) {
            mWeakActivity = new WeakReference<>(activity);
        }

        @Override
        public void run() {
            AppCompatActivity activity = mWeakActivity.get();
            if (activity != null) activity.finish();
        }

    }

}

Answer 2

您可以使用android.os.Handler类来执行此操作，像

private Handler handler = new Handler(); // Create Handler

Runnable runnable = new Runnable() {
     @Override
     public void run() {
         // Perform action here...
     }           
};
handler.postDelayed(runnable, 3 * 1000); // action will be performed after 3 seconds.

Answer 3

 CountDownTimer timer = new CountDownTimer(30000/*modify value as per need*/, 1000) {

     public void onTick(long millisUntilFinished) {
        //millisUntilFinised is the remaining time
     }

     public void onFinish() {
        //timer finished .Do what you need to do next here
     }
  };

使用timer.start();，你必须启动计时器。

如何在一定时间后关闭窗口/活动

3 个答案: