I have an Activity with a ViewPager, each page of the ViewPager has a Fragment.
Inside my Screen3Fragment I have a LinearLayout (lly_fragments) where I am showing some other fragments. I start by showing the fragment Screen3_1
public class Screen3Fragment extends Fragment {
private FragmentManager manager;
private FragmentTransaction transaction;
public static Screen3Fragment newInstance() {
final Screen3Fragment mf = new Screen3Fragment();
return mf;
}
public Screen3Fragment() {
}
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
View v = inflater.inflate(R.layout.fragment_screen3, container, false);
Screen3_1Fragment frag31 = new Screen3_1Fragment();
manager = getChildFragmentManager();
transaction = manager.beginTransaction();
transaction.add(R.id.lly_fragments,frag31,"frag31");
transaction.addToBackStack("frag31");
transaction.commit();
return v;
}
}
This works fine without problems. Problem comes when, from within frag31 (which is inside Screen3Fragment), I want to call fragt32, for that I do the following.
public class Screen3_1Fragment extends Fragment {
private ImageButton imgbt_timer;
@Nullable
@Override
public View onCreateView(LayoutInflater inflater, @Nullable ViewGroup container, @Nullable Bundle savedInstanceState) {
View v = inflater.inflate(R.layout.fragment_screen3_1,container,false);
imgbt_timer = (ImageButton) v.findViewById(R.id.bT_scr31_timer);
imgbt_timer.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
getChildFragmentManager().beginTransaction()
.replace(R.id.lly_fragments, new Screen3_2Fragment(), "frag32")
.commit();
}
});
return v;
}
}
As I read in other answers, the line transaction.replaceshould do the trick and replace the existing frag31 by the given frag32 inside the same given container lly_fragments.
However, I get java.lang.IllegalArgumentException: No view found for id..... I am not sure why.