在Sudheesh B Nair的回答中有一些改进，我注意到它会等待处理程序，即使在立即按回两次，所以取消处理程序如下所示。我已经取消吐司也防止它显示后应用程序退出。

 boolean doubleBackToExitPressedOnce = false;
        Handler myHandler;
        Runnable myRunnable;
        Toast myToast;

    @Override
        public void onBackPressed() {
            if (doubleBackToExitPressedOnce) {
                myHandler.removeCallbacks(myRunnable);
                myToast.cancel();
                super.onBackPressed();
                return;
            }

            this.doubleBackToExitPressedOnce = true;
            myToast = Toast.makeText(this, "Please click BACK again to exit", Toast.LENGTH_SHORT);
            myToast.show();

            myHandler = new Handler();

            myRunnable = new Runnable() {

                @Override
                public void run() {
                    doubleBackToExitPressedOnce = false;
                }
            };
            myHandler.postDelayed(myRunnable, 2000);
        }

以下是我的看法:

 int oddeven = 0;
 long backBtnPressed1;
 long backBtnPressed2;
 @Override
 public void onBackPressed() {
     oddeven++;
     if(oddeven%2==0){
         backBtnPressed2 = System.currentTimeMillis();
         if(backBtnPressed2-backBtnPressed1<2000) {
            super.onBackPressed();
            return;
         }
     }
     else if(oddeven%2==1) { 
         backBtnPressed1 = System.currentTimeMillis();    
        //  Insert toast back button here
     }
 }

Sudheesh B Nair对这个问题有一个很好的(被接受的)答案，我认为应该有一个更好的选择，比如;

测量所经过的时间并检查自上次回按以来TIME_INTERVAL毫秒(例如2000)是否经过了什么错误?下面的示例代码使用System.currentTimeMillis();onBackPressed()被调用来存储时间;

private static final int TIME_INTERVAL = 2000; // # milliseconds, desired time passed between two back presses.
private long mBackPressed;

@Override
public void onBackPressed()
{
    if (mBackPressed + TIME_INTERVAL > System.currentTimeMillis()) 
    { 
        super.onBackPressed(); 
        return;
    }
    else { Toast.makeText(getBaseContext(), "Tap back button in order to exit", Toast.LENGTH_SHORT).show(); }

    mBackPressed = System.currentTimeMillis();
}

回到公认的答案批评;使用一个标志来指示它是否在最后的TIME_INTERVAL(比如2000)毫秒内被按下，而set - reset是通过Handler的postDelayed()方法来执行的，这是我想到的第一件事。但是postDelayed()操作应该在活动关闭时取消，删除Runnable。

为了移除Runnable，它不能被声明为匿名，并且必须与Handler一起声明为成员。然后可以适当地调用Handler的removeCallbacks()方法。

下面的示例是演示;

private boolean doubleBackToExitPressedOnce;
private Handler mHandler = new Handler();

private final Runnable mRunnable = new Runnable() {
    @Override
    public void run() {
        doubleBackToExitPressedOnce = false;                       
    }
};

@Override 
protected void onDestroy() 
{ 
    super.onDestroy();

    if (mHandler != null) { mHandler.removeCallbacks(mRunnable); }
}

@Override
public void onBackPressed() {
    if (doubleBackToExitPressedOnce) {
        super.onBackPressed();
        return;
    }

    this.doubleBackToExitPressedOnce = true;
    Toast.makeText(this, "Please click BACK again to exit", Toast.LENGTH_SHORT).show();

    mHandler.postDelayed(mRunnable, 2000);
}

感谢@NSouth的贡献;为了防止在应用程序关闭后出现toast消息，toast可以被声明为成员，比如mExitToast，并且可以通过mExitToast.cancel()取消;就在super.onBackPressed()之前;调用。

退出应用程序时使用Runnable并不是一个好主意，我最近想出了一个更简单的方法来记录和比较两次后退按钮点击之间的时间。示例代码如下:

private static long back_pressed_time;
private static long PERIOD = 2000;

@Override
public void onBackPressed()
{
        if (back_pressed_time + PERIOD > System.currentTimeMillis()) super.onBackPressed();
        else Toast.makeText(getBaseContext(), "Press once again to exit!", Toast.LENGTH_SHORT).show();
        back_pressed_time = System.currentTimeMillis();
}

这将在特定的延迟时间内(样本为2000毫秒)单击两次BACK按钮来退出应用程序。

在java中

private Boolean exit = false; 

if (exit) {
onBackPressed(); 
}

 @Override
public void onBackPressed() {
    if (exit) {
        finish(); // finish activity
    } else {
        Toast.makeText(this, "Press Back again to Exit.",
                Toast.LENGTH_SHORT).show();
        exit = true;
        new Handler().postDelayed(new Runnable() {
            @Override
            public void run() {
                exit = false;
            }
        }, 3 * 1000);

    }
}

在kotlin

 private var exit = false

 if (exit) {
        onBackPressed()
         }

 override fun onBackPressed(){
           if (exit){
               finish() // finish activity
           }else{
            Toast.makeText(this, "Press Back again to Exit.",
                    Toast.LENGTH_SHORT).show()
            exit = true
            Handler().postDelayed({ exit = false }, 3 * 1000)

        }
    }

当你将之前的堆栈活动存储在堆栈中时，这也会有所帮助。

我修改了Sudheesh的答案

boolean doubleBackToExitPressedOnce = false;

@Override
public void onBackPressed() {
    if (doubleBackToExitPressedOnce) {
        //super.onBackPressed();

  Intent intent = new Intent(Intent.ACTION_MAIN);
                    intent.addCategory(Intent.CATEGORY_HOME);
                    intent.setFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);//***Change Here***
                    startActivity(intent);
                    finish();
                    System.exit(0);
        return;
    }

    this.doubleBackToExitPressedOnce = true;
    Toast.makeText(this, "Please click BACK again to exit", Toast.LENGTH_SHORT).show();

    new Handler().postDelayed(new Runnable() {

        @Override
        public void run() {
            doubleBackToExitPressedOnce=false;                       
        }
    }, 2000);
}