Using onbackpressed on homepage to exit app, in other case go to previous page

Question

I'm developing an Android Webview app at the moment. I'm using the following code to exit my app. The problem is that by using this code, I can't return to the previous page anymore so it's not yet what I had in mind.

Goal:

I'd like to show the below toast message only when pressing the back-button on the homepage. When the user is on a different page than the homepage, pressing the back-button should just return to the previous page.

How could I do something like that?

private Boolean exit = false;
@Override
    public void onBackPressed() {
        if (exit)
            this.finish();
        else {
            Toast.makeText(this, "Press again to close the app.",
                    Toast.LENGTH_SHORT).show();
            exit = true;
            new Handler().postDelayed(new Runnable() {
                @Override
                public void run() {
                    exit = false;
                }
            }, 3 * 1000);

        }

    }

score 0 · Answer 1 · answered Nov 30 '14 at 18:38

0

You have to have some kind of variable for knowing in which page are you.

private Boolean exit = false;
@Override
public void onBackPressed() {
     if(page != "homepage"){
          super.onBackPressed(); // Calls the Overriden Method 
     }
    else
    {
    if (exit)
        this.finish();
    else {
        Toast.makeText(this, "Press again to close the app.",
                Toast.LENGTH_SHORT).show();
        exit = true;
        new Handler().postDelayed(new Runnable() {
            @Override
            public void run() {
                exit = false;
            }
        }, 3 * 1000);

    }
  }

}

answered Nov 30 '14 at 18:38

PaytoN

154
10

Thanks for the reply! I had something like that in mind, but my real problem is that I don't exactly get how to define that variable? Let's say my website is called http://example.com , how should I define the page variable? Because right now I'm getting the error "page cannot be resolved to a variable" (which is normal). – Stan Nov 30 '14 at 19:01
You can split your URL ( http://yoururl.com/index.php ) and taking the name of 'index.php' or '' – PaytoN Nov 30 '14 at 19:48

score 0 · Answer 2 · answered Nov 30 '14 at 18:55

If you are working with activities then try to declare their parent activities in AndroidManifest.XML
It will also create the backButton on the actionBar.

  <activity
        android:name=".MyActivity"
        android:label="@string/app_name"
        android:parentActivityName="com.projectname.homepage">
        <meta-data
            android:name="android.support.PARENT_ACTIVITY"
            android:value="com.projectname.homepage" />
    </activity>

In your Homepage declare this code onBackPressed.

@Override
public void onBackPressed() {
    if (back_pressed + 2000 > System.currentTimeMillis()) {
        Intent intent = new Intent();
        intent.setFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
        super.onBackPressed();
    } else
    Toast.makeText(this, "Press once again to exit!",
            Toast.LENGTH_SHORT).show();
    back_pressed = System.currentTimeMillis();
}

Using onbackpressed on homepage to exit app, in other case go to previous page

2 Answers2