Clicking the back button twice to exit an activity

Question

I've noticed this pattern in a lot of Android apps and games recently: when clicking the back button to "exit" the application, a Toast comes up with a message similar to "Please click BACK again to exit".

I was wondering, as I'm seeing it more and more often, is that a built-in feature that you can somehow access in an activity? I've looked at the source code of many classes but I can't seem to find anything about that.

Of course, I can think about a few ways to achieve the same functionality quite easily (the easiest is probably to keep a boolean in the activity that indicates whether the user already clicked once...) but I was wondering if there's something already here.

EDIT: As @LAS_VEGAS mentioned, I didn't really mean "exit" in the traditional meaning. (i.e. terminated) I meant "going back to whatever was open before the application start activity was launched", if that makes sense :)

Answer 1

@Override public void onBackPressed() {
   Log.d("CDA", "onBackPressed Called");
   Intent intent = new Intent();
   intent.setAction(Intent.ACTION_MAIN);
   intent.addCategory(Intent.CATEGORY_HOME);

   startActivity(intent);
}

Answer 2

 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().postDelayed(new Runnable() {

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

Declare Variableprivate boolean doubleBackToExitPressedOnce = false;

Paste this in your Main Activity and this will solve your issue

Answer 3

The Accepted answer is Best one but if you are using Android Design Support Library then you can use SnackBar for Better Views.

   boolean doubleBackToExitPressedOnce = false;

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

        this.doubleBackToExitPressedOnce = true;

        Snackbar.make(findViewById(R.id.photo_album_parent_view), "Please click BACK again to exit", Snackbar.LENGTH_SHORT).show();

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

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

Answer 4

There is very simplest way among all these answers.

Simply write following code inside onBackPressed() method.

long back_pressed;

@Override
public void onBackPressed() {
    if (back_pressed + 1000 > System.currentTimeMillis()){
        super.onBackPressed();
    }
    else{
        Toast.makeText(getBaseContext(),
                "Press once again to exit!", Toast.LENGTH_SHORT)
                .show();
    }
    back_pressed = System.currentTimeMillis();
}

You need to define back_pressed object as long in activity.

Answer 5

It is not a good idea to use a Runnable when exit the application, I recently figure out a much simpler way to record and compare the period between two BACK button clicks. Sample code as following:

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();
}

This will do the trick to exit the application by a double BACK button clicks within a certain delay period which is 2000 millisecond in sample.

Answer 6

It's not a built in functionality. I think it is not even the recommended behavior. Android apps are not meant to exit:

Why dont Android applications provide an "Exit" option?