RecyclerView: IndexOutOfBoundsException. Invalid item position

Question

I have two fragments: A & B.

Fragment B contains SearchView and RecyclerView.

adapter = new SearchListAdapter(items);

recyclerView = new RecyclerView(getContext());
recyclerView.setHasFixedSize(true);
recyclerView.setAdapter(adapter);
recyclerView.setLayoutManager(new LinearLayoutManager(getContext()));

params = new RelativeLayout.LayoutParams(LayoutParams.MATCH_PARENT, LayoutParams.MATCH_PARENT);
params.addRule(BELOW, SEARCH_VIEW_ID);
addView(recyclerView, params);

@Override
public boolean onQueryTextChange(String newText)
{
    ArrayList<MyModel> filteredModelList = adapter.filter(newText);
    recyclerView.scrollToPosition(0);

    return true;
}

Implementing filtering is taken away: How to filter a RecyclerView with a SearchView

PROBLEM

1) At the beginning we are on a fragment A, then go to the fragment B is administered in SearchView more characters and get a filtered list (for example consisting of a single element).

2) Next, through getFragmentManager().popBackStack(); we return to fragment A.

3) At the same time, in the fragment B in the method of onStop(); we clean SearchView:

@Override
public void onStop()
{
    if(searchView != null)
    {
        searchView.setQuery("", false);
    }
}

4) Next, we return to the fragment B and get an error:

java.lang.IndexOutOfBoundsException: Inconsistency detected. Invalid item position 1(offset:35).state:35

I've been looking at this problem and found the following:

a) Issue 77846

b) Issue 77232

c) habra

d) recyclerview layoutmanager part 1 (generateDefaultLayoutParams)

e) IndexOutOfBoundsException Invalid item position XX(XX). Item count:XX

He came to the conclusion that this is a bug recyclerview and it has not yet been fixed.

As a result, it was necessary to implement filtering without animation:

public void filter(String charText)
{
    charText = charText.toLowerCase();
    mItems = new ArrayList<MyModel>();
    if (charText.length() == 0)
    {
        mItems.addAll(mOriginalItems);
    }
    else
    {
        for (MyModel model : mOriginalItems)
        {
            String textId = model.getId().toLowerCase();
            if (textId.contains(charText))
            {
                mItems.add(model);
            }
        }
    }

    notifyDataSetChanged();
}

Is this true, and there is an error in the code?

Thank you!

UPDATE

Activity:

public class MyActivity extends AppCompatActivity

SimpleFragmentManager:

private static void moveTo(BaseFragment fragment)
{
    if(fragment != null)
    {
        FragmentManager fragmentManager = _activity.getFragmentManager();
        FragmentTransaction ft = fragmentManager.beginTransaction();
        BaseFragment currentFragment = getCurrentFragment();

        if(fragment != currentFragment)
        {
            ft.replace(_containerId, fragment, fragment.getType());
            if (currentFragment != null) ft.addToBackStack(null);
            ft.commit();
        }
    }
}

Fragment A & B:

@Nullable
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState)
{
    if(view == null)
    {
        view = new ALayout(container.getContext());
        //if fragment B then > view = new BLayout(container.getContext());
        view.setLayoutParams(new RelativeLayout.LayoutParams(RelativeLayout.LayoutParams.MATCH_PARENT, RelativeLayout.LayoutParams.MATCH_PARENT));
    }

    return view;
}

ALayout & BLayout is extends RelativeLayout

BLayout with SearchView and RecycleView I showed above.

Important Notice:

1) if comment out line: recyclerView.scrollToPosition(0); - no bug! (and no scroll top :) )

2) More precise conditions for obtaining bug:

a) Go to the fragment_1 track on the fragment_2;

b) Enter a query in the search (String is present: recyclerView.scrollToPosition(0));

c) Return to fragment_1 (this time a search query to clean - no errors!);

d) We will again go to the fragment_2.

>>We get a error!

a) Go to the fragment_1 track on the fragment_2;

b) Do not enter anything in the search (String is present: recyclerView.scrollToPosition (0));

c) Go back to the fragment_1. (this time a search query to clean - no errors!);

d) Once again we are going to fragment_2.

>>No error!