java - 菜单打开时覆盖后退按钮

Question

我想要实现的是，当用户在 MENU 可见时单击后退按钮时，menuActual 和 MENU 的状态从可见变为不可见。如果 MENU 未打开并且用户单击返回，则会显示一个 Toast，并显示“再次按下即可退出”，并且如果您单击2 秒内返回，应用程序将关闭。

我拥有的代码:

@Override
    public void finish() {

    if (MENU.getVisibility() == View.VISIBLE){
        MENU.setVisibility(View.INVISIBLE);
        menuActual.setVisibility(View.INVISIBLE);

    }else {
        if (doubleBackToExitPressedOnce) {
            super.onBackPressed();
            moveTaskToBack(true);

            return;
        }else {

            this.doubleBackToExitPressedOnce = true;
            Toast.makeText(this, "Tap again to Exit!", Toast.LENGTH_SHORT).show();

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

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

我已声明 boolean doubleBackToExitPressedOnce = false;

应用程序甚至显示 Toast 说“再次按退出”，但如果再次单击返回，应用程序会说“AppName isn' t 响应”

努力找出原因，这是漫长的一天。

谢谢!

Answer 1

在onBackPressed中这样做:

private boolean doubleBackToExitPressedOnce = false;
private Handler handler;
private Runnable runnable;

@Override
public void onBackPressed() {
    if (MENU.getVisibility() == View.VISIBLE) {
        MENU.setVisibility(View.INVISIBLE);
        menuActual.setVisibility(View.INVISIBLE);
        return;
    }

    if (!doubleBackToExitPressedOnce) {
        doubleBackToExitPressedOnce = true;
        Toast.makeText(this, "Tap again to Exit!", Toast.LENGTH_SHORT).show();

        handler = new Handler();
        handler.postDelayed(runnable = new Runnable() {

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

    // Removes the callBack
    handler.removeCallbacks(runnable);

    // Replace this next line with finishAffinity() if you want to close the app.
    super.onBackPressed();
}