DialogFragment not found in fragmentTransaction

Question

I am displaying DialogFragment from a manager. DialogFragment display multiple times.

I want to know is there a way to check from transaction whether this fragment already displaying. So don't display it.

@Override
public void show(FragmentManager manager, String tag) {
    try {
        FragmentTransaction ft = manager.beginTransaction();

        Fragment prev = manager.findFragmentByTag(tag);
        if (prev == null) {
            ft.add(this, tag);
            ///ft.addToBackStack(tag);
            ft.commitAllowingStateLoss();
        }

    } catch (IllegalStateException e) {
        Log.d("ABSDIALOGFRAG", "Exception", e);
    }
}

I am calling my Fragment like

CustomerFeedbackDialog feedbackDialog = CustomerFeedbackDialog.newInstance(genaric.getData(), type);

feedbackDialog.show(getSupportFragmentManager(), "feedbackDialog");

I have call findFragmentByTag but it is always null. I don't want to show already displayed Fragment. otherwise it duplicate . Mulitple dialogFragment opens

I know I can do it using a flag in sharedprefs

EDIT Solution found

Thanks for your help. Problem solved and I posted answer below

Answer 1

I finally able to handle it by overriding show method and addToBackStack(null) and executePendingTransactions

Firstly to put tag in findFragmentByTag addToBackStack is must otherwise tag is not added in fragmentTransaction.

    @Override
    public void show(FragmentManager manager, String tag) {
        try {
            FragmentTransaction ft = manager.beginTransaction();

            Fragment prev = manager.findFragmentByTag(tag);
            if (prev == null) {
                ft.add(this, tag);
                ft.addToBackStack(null);
                ft.commitAllowingStateLoss();
                manager.executePendingTransactions();
            }

        } catch (IllegalStateException e) {
            Log.d("ABSDIALOGFRAG", "Exception", e);
        }
    }

Now if fragment is already in transaction. It will not display again..