最近我在许多Android应用和游戏中注意到这种模式:当点击后退按钮“退出”应用时,Toast会出现类似于“请再次点击后退退出”的消息。

我在想,当我越来越频繁地看到它时,这是一个内置的功能,你可以在某个活动中访问它吗?我已经看了很多类的源代码,但我似乎找不到任何关于这一点。

当然,我可以想到一些很容易实现相同功能的方法(最简单的可能是在活动中保留一个布尔值,指示用户是否已经单击过一次…),但我想知道这里是否已经有一些东西。

编辑:正如@LAS_VEGAS所提到的,我并不是指传统意义上的“退出”。(即终止)我的意思是“回到应用程序启动活动启动之前打开的任何东西”,如果这有意义的话:)


当前回答

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);
        }

其他回答

当HomeActivity包含导航抽屉和双backPressed()函数退出应用程序。 (不要忘记初始化全局变量布尔doubleBackToExitPressedOnce = false;) 将doubleBackPressedOnce变量设置为false

@Override
public void onBackPressed() {
    DrawerLayout drawer = findViewById(R.id.drawer_layout);
    if (drawer.isDrawerOpen(GravityCompat.END)) {
        drawer.closeDrawer(GravityCompat.END);
    } else {
        if (doubleBackToExitPressedOnce) {
            super.onBackPressed();
            moveTaskToBack(true);
            return;
        } else {
            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);
        }
    }
}

在Java活动中:

boolean doubleBackToExitPressedOnce = false;

@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();
        
    new Handler(Looper.getMainLooper()).postDelayed(new Runnable() {
        
        @Override
        public void run() {
            doubleBackToExitPressedOnce=false;                       
        }
    }, 2000);
} 

在Kotlin活动:

private var doubleBackToExitPressedOnce = false
override fun onBackPressed() {
        if (doubleBackToExitPressedOnce) {
            super.onBackPressed()
            return
        }

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

        Handler(Looper.getMainLooper()).postDelayed(Runnable { doubleBackToExitPressedOnce = false }, 2000)
    }

我认为这个处理程序有助于在2秒后重置变量。

我认为这个方法比Zefnus好一点。只调用一次System.currentTimeMillis()并忽略return;:

long previousTime;

@Override
public void onBackPressed()
{
    if (2000 + previousTime > (previousTime = System.currentTimeMillis())) 
    { 
        super.onBackPressed();
    } else {
        Toast.makeText(getBaseContext(), "Tap back button in order to exit", Toast.LENGTH_SHORT).show();
    }
}

按键2次返回

public void click(View view){
    if (isBackActivated) {
        this.finish();
    }
    if (!isBackActivated) {
        isBackActivated = true;
        Toast.makeText(getApplicationContext(), "Again", Toast.LENGTH_SHORT).show();
        Handler handler = new Handler();
        handler.postDelayed(new Runnable() {
            @Override
            public void run() {
                isBackActivated = false;  // setting isBackActivated after 2 second
            }
        }, 2000);
    }

}

在Kotlin的背面按下退出应用程序,你可以使用:

定义一个全局变量:

private var doubleBackToExitPressedOnce = false

覆盖onBackPressed:

override fun onBackPressed() {
        if (doubleBackToExitPressedOnce) {
            super.onBackPressed()
            return
        }

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

        Handler().postDelayed({
            doubleBackToExitPressedOnce = false;
        }, 2000)
    }