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

在这种情况下,Snackbar是比Toast更好的选择来显示退出动作。下面是小吃店的工作方法。

@Override
        public void onBackPressed() {
            if (doubleBackToExitPressedOnce) {
                super.onBackPressed();
                return;
            }
            this.doubleBackToExitPressedOnce = true;
            Snackbar.make(this.getWindow().getDecorView().findViewById(android.R.id.content), "Please click BACK again to exit", Snackbar.LENGTH_SHORT).show();

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

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

根据正确的答案和评论中的建议,我创建了一个演示,工作绝对很好,并在使用后删除处理程序回调。

MainActivity.java

package com.mehuljoisar.d_pressbacktwicetoexit;

import android.os.Bundle;
import android.os.Handler;
import android.app.Activity;
import android.widget.Toast;

public class MainActivity extends Activity {

    private static final long delay = 2000L;
    private boolean mRecentlyBackPressed = false;
    private Handler mExitHandler = new Handler();
    private Runnable mExitRunnable = new Runnable() {

        @Override
        public void run() {
            mRecentlyBackPressed=false;   
        }
    };

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
    }

    @Override
    public void onBackPressed() {

        //You may also add condition if (doubleBackToExitPressedOnce || fragmentManager.getBackStackEntryCount() != 0) // in case of Fragment-based add
        if (mRecentlyBackPressed) {
            mExitHandler.removeCallbacks(mExitRunnable);
            mExitHandler = null;
            super.onBackPressed();
        }
        else
        {
            mRecentlyBackPressed = true;
            Toast.makeText(this, "press again to exit", Toast.LENGTH_SHORT).show();
            mExitHandler.postDelayed(mExitRunnable, delay);
        }
    }

}

希望对大家有所帮助!!

我知道这是一个很老的问题,但这是你想做的最简单的方法。

@Override
public void onBackPressed() {
   ++k; //initialise k when you first start your activity.
   if(k==1){
      //do whatever you want to do on first click for example:
      Toast.makeText(this, "Press back one more time to exit", Toast.LENGTH_LONG).show();
   }else{
      //do whatever you want to do on the click after the first for example:
      finish(); 
   }
}

我知道这不是最好的方法,但它很有效!

下面是一种使用RxJava的方法:

override fun onCreate(...) {
    backPresses.timeInterval(TimeUnit.MILLISECONDS, Schedulers.io())
            .skip(1) //Skip initial event; delay will be 0.
            .onMain()
            .subscribe {
                if (it.time() < 7000) super.onBackPressed() //7000 is the duration of a Toast with length LENGTH_LONG.
            }.addTo(compositeDisposable)

    backPresses.throttleFirst(7000, TimeUnit.MILLISECONDS, Schedulers.io())
            .subscribe { Toast.makeText(this, "Press back again to exit.", LENGTH_LONG).show() }
            .addTo(compositeDisposable)
}

override fun onBackPressed() = backPresses.onNext(Unit)

下面是完整的工作代码。并且不要忘记删除回调,这样它就不会在应用程序中导致内存泄漏:)

private boolean backPressedOnce = false;
private Handler statusUpdateHandler;
private Runnable statusUpdateRunnable;

public void onBackPressed() {
        if (backPressedOnce) {
            finish();
        }

        backPressedOnce = true;
        final Toast toast = Toast.makeText(this, "Press again to exit", Toast.LENGTH_SHORT);
        toast.show();

        statusUpdateRunnable = new Runnable() {
            @Override
            public void run() {
                backPressedOnce = false;
                toast.cancel();  //Removes the toast after the exit.
            }
        };

        statusUpdateHandler.postDelayed(statusUpdateRunnable, 2000);
}

@Override
protected void onDestroy() {
    super.onDestroy();
    if (statusUpdateHandler != null) {
        statusUpdateHandler.removeCallbacks(statusUpdateRunnable);
    }
}