Android ViewPager get the current View

Question

I have a ViewPager, and I'd like to get the current selected and visible view, not a position.

1. getChildAt(getCurrentItem) returns wrong View

2. This works not all the time. Sometimes returns null, sometimes just returns wrong View.

   @Override
   public void setUserVisibleHint(boolean isVisibleToUser) {
       super.setUserVisibleHint(isVisibleToUser);
   
       if (isVisibleToUser == true) { 
           mFocusedListView = ListView; 
       }
   }
   

3. PageListener on ViewPager with getChildAt() also not working, not giving me the correct View every time.

How can i get current visible View?

View view = MyActivity.mViewPager.getChildAt(MyActivity.mViewPager.getCurrentItem()).getRootView();
ListView listview = (ListView) view.findViewById(R.id.ListViewItems);

Answer 1

I've figured it out. What I did was to call setTag() with a name to all Views/ListViews, and just call findViewWithTag(mytag), mytag being the tag.

Unfortunately, there's no other way to solve this.

Answer 2

I just came across the same issue and resolved it by using:

View view = MyActivity.mViewPager.getFocusedChild();

Answer 3

I use this method with android.support.v4.view.ViewPager

View getCurrentView(ViewPager viewPager) {
        try {
            final int currentItem = viewPager.getCurrentItem();
            for (int i = 0; i < viewPager.getChildCount(); i++) {
                final View child = viewPager.getChildAt(i);
                final ViewPager.LayoutParams layoutParams = (ViewPager.LayoutParams) child.getLayoutParams();

                Field f = layoutParams.getClass().getDeclaredField("position"); //NoSuchFieldException
                f.setAccessible(true);
                int position = (Integer) f.get(layoutParams); //IllegalAccessException

                if (!layoutParams.isDecor && currentItem == position) {
                    return child;
                }
            }
        } catch (NoSuchFieldException e) {
            Log.e(TAG, e.toString());
        } catch (IllegalArgumentException e) {
            Log.e(TAG, e.toString());
        } catch (IllegalAccessException e) {
            Log.e(TAG, e.toString());
        }
        return null;
    }

Answer 4

You can get the current element by accessing your list of itens from your adapter calling myAdapter.yourListItens.get(myViewPager.getCurrentItem()); As you can see, ViewPager can retrieve the current index of element of you adapter (current page).

If you is using FragmentPagerAdapter you can do this cast:

FragmentPagerAdapter adapter = (FragmentPagerAdapter)myViewPager.getAdapter();

and call

adapter.getItem(myViewPager.getCurrentItem());

This works very well for me ;)

Answer 5

During my endeavors to find a way to decorate android views I think I defined alternative solution for th OP's problem that I have documented in my blog. I am linking to it as the code seems to be a little bit too much for including everything here.

The solution I propose:

• keeps the adapter and the view entirely separated
• one can easily query for a view with any index form the view pager and he will be returned either null if this view is currently not loaded or the corresponding view.

Answer 6

Use an Adapter extending PagerAdapter, and override setPrimaryItem method inside your PagerAdapter.

https://developer.android.com/reference/android/support/v4/view/PagerAdapter.html

class yourPagerAdapter extends PagerAdapter
{
    // .......

    @Override
    public void setPrimaryItem (ViewGroup container, int position, Object object)
    {
        int currentItemOnScreenPosition = position;
        View onScreenView = getChildAt(position);
    }

    // .......

}

Answer 7

viewpager.getChildAt(0)

this always returns my currently selected view. this worked for me.

Answer 8

Try this

 final int position = mViewPager.getCurrentItem();
    Fragment fragment = getSupportFragmentManager().findFragmentByTag("android:switcher:" + R.id.rewards_viewpager + ":"
            + position);

Answer 9

I had to do it more general, so I decided to use the private 'position' of ViewPager.LayoutParams

        final int childCount = viewPager.getChildCount();
        for (int i = 0; i < childCount; i++) {
            final View child = viewPager.getChildAt(i);
            final ViewPager.LayoutParams lp = (ViewPager.LayoutParams) child.getLayoutParams();
            int position = 0;
            try {
                Field f = lp.getClass().getDeclaredField("position");
                f.setAccessible(true);
                position = f.getInt(lp); //IllegalAccessException
            } catch (NoSuchFieldException | IllegalAccessException ex) {ex.printStackTrace();}
            if (position == viewPager.getCurrentItem()) {
                viewToDraw = child;
            }
        }

Answer 10

I'm using ViewPagerUtils from FabulousFilter:

ViewPagerUtils.getCurrentView(ViewPager viewPager)

Answer 11

If you do not have many pages and you can safely apply setOffscreenPageLimit(N-1) where N is the total number of pages without wasting too much memory then you could do the following:

public Object instantiateItem(final ViewGroup container, final int position) {      
    CustomHomeView RL = new CustomHomeView(context);
    if (position==0){
        container.setId(R.id.home_container);} ...rest of code

then here is code to access your page

((ViewGroup)pager.findViewById(R.id.home_container)).getChildAt(pager.getCurrentItem()).setBackgroundColor(Color.BLUE);

If you want you can set up a method for accessing a page

RelativeLayout getPageAt(int index){
    RelativeLayout rl =  ((RelativeLayout)((ViewGroup)pager.findViewById(R.id.home_container)).getChildAt(index));
    return rl;
}