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

 private static final int TIME_DELAY = 2000;
    private static long back_pressed;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
    }
    @Override
    public void onBackPressed() {
        if (back_pressed + TIME_DELAY > System.currentTimeMillis()) {
            super.onBackPressed();
        } else {
            Toast.makeText(getBaseContext(), "Press once again to exit!",
                    Toast.LENGTH_SHORT).show();
        }
        back_pressed = System.currentTimeMillis();
    }

其他回答

我用这个

import android.app.Activity;
import android.support.annotation.StringRes;
import android.widget.Toast;

public class ExitApp {

    private static long lastClickTime;

    public static void now(Activity ctx, @StringRes int message) {
        now(ctx, ctx.getString(message), 2500);
    }

    public static void now(Activity ctx, @StringRes int message, long time) {
        now(ctx, ctx.getString(message), time);
    }

    public static void now(Activity ctx, String message, long time) {
        if (ctx != null && !message.isEmpty() && time != 0) {
            if (lastClickTime + time > System.currentTimeMillis()) {
                ctx.finish();
            } else {
                Toast.makeText(ctx, message, Toast.LENGTH_SHORT).show();
                lastClickTime = System.currentTimeMillis();
            }
        }
    }

}

使用到事件onBackPressed

@Override
public void onBackPressed() {
   ExitApp.now(this,"Press again for close");
}

或ExitApp.now(这个,R.string.double_back_pressed)

对于需要关闭的更改秒,指定毫秒

ExitApp.now (R.string.double_back_pressed, 5000)

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)
    }

这个解的独特之处在于它的行为;其中,非双击将显示吐司和成功双击将显示没有吐司,同时关闭应用程序。 唯一的缺点是吐司的显示将有650毫秒的延迟。我相信这是最佳行为的最佳解决方案,因为逻辑表明,如果没有这样的延迟,就不可能有这种行为

//App Closing Vars
private var doubleBackPressedInterval: Long = 650
private var doubleTap = false
private var pressCount = 0
private var timeLimit: Long = 0

override fun onBackPressed() {
    pressCount++
    if(pressCount == 1) {
        timeLimit = System.currentTimeMillis() + doubleBackPressedInterval
        if(!doubleTap) {
            showExitInstructions()
        }
    }
    if(pressCount == 2) {
        if(timeLimit > System.currentTimeMillis()) {
            doubleTap = true
            super.onBackPressed()
        }
        else {
            showExitInstructions()
        }
        pressCount = 1
        timeLimit = System.currentTimeMillis() + doubleBackPressedInterval
    }
}

private fun showExitInstructions() {
    Handler().postDelayed({
        if(!doubleTap) {
            Toast.makeText(this, "Try Agian", Toast.LENGTH_SHORT).show()
        }
    }, doubleBackPressedInterval)
}
private static final int TIME_INTERVAL = 2000;
private long mBackPressed;
    @Override
        public void onBackPressed() {

            if (mBackPressed + TIME_INTERVAL > System.currentTimeMillis()) {
                super.onBackPressed();
                Intent intent = new Intent(FirstpageActivity.this,
                        HomepageActivity.class);
                startActivity(intent);
                finish();

                return;
            } else {

                Toast.makeText(getBaseContext(),
                        "Tap back button twice  to go Home.", Toast.LENGTH_SHORT)
                        .show();

                mBackPressed = System.currentTimeMillis();

            }

        }