Android ViewPager get the current View

问题:

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);

回答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.

回答2:

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 ;)

回答3:

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

View view = MyActivity.mViewPager.getFocusedChild();

回答4:

how does this play out for all of y'all?

public Object instantiateItem(final ViewGroup container, final int position) {           CustomHomeView RL = new CustomHomeView(MainActivity.ctxt);     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 little meathod

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

回答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.

回答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;     }      // .......  }

回答7:

is that your first activity on the screen or have you layered some above each other already?

try this:

findViewById(android.R.id.content).getRootView()

or just:

findViewById(android.R.id.content) 

also depending on what you want try:

((ViewGroup)findViewById(android.R.id.content)).getChildAt(0)

回答8:

Try this

 final int position = mViewPager.getCurrentItem();     Fragment fragment = getSupportFragmentManager