Android Find ListView instance which is created by Class in View Pager

Question

I have to create one reporting application, in which sane report format, adapter is used to generate reports layouts dynamically in view pager adapter.

 switch (index) {
    case 0:
        return new sfg();
    case 1:
        return  new Report1("1STREPORT");
    case 2:
        return new Report1("2NDREPORT");
    case 3:
        return new Report1("3RDREPORT");
    case 4:
        // Movies fragment activity
        return new Report1("4THREPORT");
    case 5:
        return new Report1("5THREPORT");
    case 6:
        return new Report1("Numbers");
    case 7:
        return new report2("6THREPORT");
    default:
        return  null;
}

In Report1 layout page contains two things, one Button and another is List View.

All the reports are displayed simultaneously.

I have to change the ListView item when I click on Button. I tried this:

public void btn_DetailedReport_Show(View view) {

    FragmentActivity myActivity= (FragmentActivity)(view.getContext());

    ListView lst1  = (ListView) view.findViewById(R.id.lst_abc);  // lst_abc is the list box Name

but lst1 is not able to call this ListView. I tried with Tag also as suggested in one example.

How can I solve this issue?

score 0 · Answer 1 · answered Jul 04 '15 at 05:29

If you want to access listView in fragment then it should be done like this...

In your onCreateView

View rootView = inflater.inflate(R.layout.fragment_name, container, false);
ListView lst1 = (ListView)rootView.findViewById(R.id.lst_abc);
return rootView;

you can declare ListView object as a global variable and then you can access it.

Reason behind this is you are not looking listView on you fragment you are looking in you button's view.

Android Find ListView instance which is created by Class in View Pager

1 Answers1