Question

I have an application which requires the user to log in to see the content. When the user pushes the back button, he is given a prompt if he wants to log out, or stay on the application. However, I want to make it so that if the user presses the back button again, the application sends him back to the log out screen. How can I do this?

@Override
public void onBackPressed() {
    exitPrompt();

}
private void exitUser(){
    UserFunctions userFunk = new UserFunctions();   
    userFunk.logoutUser(getApplicationContext());
    userFunk.resetCarcoord(getApplicationContext());
    Intent logout = new Intent(getApplicationContext(), MainActivity.class);
    logout.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
    startActivity(logout);
    finish();

}
private void exitPrompt(){
    new AlertDialog.Builder(this)
    .setIcon(android.R.drawable.ic_dialog_alert)
    .setTitle("Log out")
    .setMessage("Are you sure you want to log out?")
    .setPositiveButton("Yes", new DialogInterface.OnClickListener() {   
        @Override
        public void onClick(DialogInterface dialog, int which) {
            // TODO Auto-generated method stub
            exitUser();
        }
    })
    .setNegativeButton("No", null)
    .show();
}

}

Answer 1

Define a class with a Boolean variable mIsexitPromptShowing and AlertDialog instance, when exitPrompt() is called; make mIsexitPromptShowing instance as true and store AlertDialog instance.

On onBackPressed() function:

public void onBackPressed() {

   if (!mIsexitPromptShowing) {
      exitPrompt();
   } else {
      AlertDialog.dismiss();
   }

}

I have just given a sample code, please do handle the safe check.
Hopes this helps.