我想，我有ViewPager的逻辑。

如果我需要刷新一组页面并基于新的数据集显示它们，我调用notifyDataSetChanged()。 然后，ViewPager对getItemPosition()进行多次调用，将Fragment作为对象传递给那里。这个片段可以来自旧数据集(我想要丢弃的数据集)，也可以来自新数据集(我想要显示的数据集)。因此，我重写了getItemPosition()，在那里我必须以某种方式确定我的片段是来自旧数据集还是来自新数据集。

In my case I have a 2-pane layout with a list of top items on the left pane and a swipe view (ViewPager) on the right. So, I store a link to my current top item inside my PagerAdapter and also inside of each instantiated page Fragment. When the selected top item in the list changes, I store the new top item in PagerAdapter and call notifyDataSetChanged(). And in the overridden getItemPosition() I compare the top item from my adapter to the top item from my fragment. And only if they are not equal, I return POSITION_NONE. Then, PagerAdapter reinstantiates all the fragments that have returned POSITION_NONE.

请注意。存储顶部项id而不是引用可能是一个更好的主意。

下面的代码片段有点示意图，但我从实际工作的代码中改编了它。

public class SomeFragment extends Fragment {
  private TopItem topItem;
}

public class SomePagerAdapter extends FragmentStatePagerAdapter {
  private TopItem topItem;

  public void changeTopItem(TopItem newTopItem) {
    topItem = newTopItem;
    notifyDataSetChanged();
  }

  @Override
  public int getItemPosition(Object object) {
    if (((SomeFragment) object).getTopItemId() != topItem.getId()) {
      return POSITION_NONE;
    }
    return super.getItemPosition(object);
  }
}

感谢之前所有的研究人员!

这是一个可怕的问题，我很乐意提出一个极好的解决方案;简单、高效、有效!

如下所示，代码显示了使用一个标志来指示何时返回POSITION_NONE

public class ViewPagerAdapter extends PagerAdapter
{
    // Members
    private boolean mForceReinstantiateItem = false;

    // This is used to overcome terrible bug that Google isn't fixing
    // We know that getItemPosition() is called right after notifyDataSetChanged()
    // Therefore, the fix is to return POSITION_NONE right after the notifyDataSetChanged() was called - but only once
    @Override
    public int getItemPosition(Object object)
    {
        if (mForceReinstantiateItem)
        {
            mForceReinstantiateItem = false;
            return POSITION_NONE;
        }
        else
        {
            return super.getItemPosition(object);
        }
    }

    public void setData(ArrayList<DisplayContent> newContent)
    {
        mDisplayContent = newContent;
        mForceReinstantiateItem = true;
        notifyDataSetChanged();
    }

}

1.首先，您必须在Pageradapter类中设置getItemposition方法 2.您必须阅读您的视图寻呼机的确切位置 3.然后将该位置作为新位置的数据位置发送 4.在setonPageChange监听器中写入更新按钮的点击监听器

我修改了程序代码，只设置了特定的位置元素

public class MyActivity extends Activity {

private ViewPager myViewPager;
private List<String> data;
public int location=0;
public Button updateButton;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    data = new ArrayList<String>();
    data.add("A");
    data.add("B");
    data.add("C");
    data.add("D");
    data.add("E");
    data.add("F");

    myViewPager = (ViewPager) findViewById(R.id.pager);
    myViewPager.setAdapter(new MyViewPagerAdapter(this, data));

      updateButton = (Button) findViewById(R.id.update);

    myViewPager.setOnPageChangeListener(new ViewPager.OnPageChangeListener() {
        @Override
        public void onPageScrolled(int i, float v, int i2) {
             //Toast.makeText(MyActivity.this, i+"  Is Selected  "+data.size(), Toast.LENGTH_SHORT).show();
        }

        @Override
        public void onPageSelected( int i) {
          // here you will get the position of selected page
            final int k = i;
             updateViewPager(k);

        }

        @Override
        public void onPageScrollStateChanged(int i) {

        }
    });
}

private void updateViewPager(final int i) {  
    updateButton.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View v) {

            Toast.makeText(MyActivity.this, i+"  Is Selected  "+data.size(), Toast.LENGTH_SHORT).show();
            data.set(i, "Replaced "+i);         
            myViewPager.getAdapter().notifyDataSetChanged();
        }
    });

}

private class MyViewPagerAdapter extends PagerAdapter {

    private List<String> data;
    private Context ctx;

    public MyViewPagerAdapter(Context ctx, List<String> data) {
        this.ctx = ctx;
        this.data = data;
    }

    @Override
    public int getCount() {
        return data.size();
    }

    @Override
    public int getItemPosition(Object object) {
        return POSITION_NONE;
    }

    @Override
    public Object instantiateItem(View collection, int position) {          

        TextView view = new TextView(ctx);
        view.setText(data.get(position));
        ((ViewPager)collection).addView(view);            
        return view;
    }

    @Override
    public void destroyItem(View collection, int position, Object view) {
         ((ViewPager) collection).removeView((View) view);
    }

    @Override
    public boolean isViewFromObject(View view, Object object) {
        return view == object;
    }

    @Override
    public Parcelable saveState() {
        return null;
    }

    @Override
    public void restoreState(Parcelable arg0, ClassLoader arg1) {
    }

    @Override
    public void startUpdate(View arg0) {
    }

    @Override
    public void finishUpdate(View arg0) {
    }
}
}

在OP提出他的问题两年半之后，这个问题仍然，嗯，仍然是一个问题。显然谷歌在这方面的优先级不是特别高，所以我没有找到解决方案，而是找到了一个变通办法。对我来说，最大的突破是找到了问题的真正原因(见本文中公认的答案)。一旦问题明显是任何活动页面都没有正确刷新，我的解决方法就很明显了:

在我的片段(几页)中:

I took all the code which populates the form out of onCreateView and put it in a function called PopulateForm which may be called from anywhere, rather than by the framework. This function attempts to get the current View using getView, and if that is null, it just returns. It's important that PopulateForm contains only the code that displays - all the other code which creates FocusChange listeners and the like is still in OnCreate Create a boolean which can be used as a flag indicating the form must be reloaded. Mine is mbReloadForm Override OnResume() to call PopulateForm() if mbReloadForm is set.

在我的活动中，我做页面的加载:

Go to page 0 before changing anything. I'm using FragmentStatePagerAdapter, so I know that two or three pages are affected at most. Changing to page 0 ensures I only ever have the problem on pages 0, 1 and 2. Before clearing the old list, take it's size(). This way you know how many pages are affected by the bug. If > 3, reduce it to 3 - if you're using a a different PagerAdapter, you'll have to see how many pages you have to deal with (maybe all?) Reload the data and call pageAdapter.notifyDataSetChanged() Now, for each of the affected pages, see if the page is active by using pager.getChildAt(i) - this tells you if you have a view. If so, call pager.PopulateView(). If not, set the ReloadForm flag.

在此之后，当您重新加载第二组页面时，该错误仍然会导致一些页面显示旧数据。但是，现在它们将被刷新，您将看到新的数据-您的用户不会知道页面是不正确的，因为这种刷新将在他们看到页面之前发生。

希望这能帮助到一些人!

你只需要这段简单的代码:

代码:

private void updateAdapter() {
    if (adapterViewPager != null) {
        int from = vpMyViewPager.getCurrentItem() - vpMyViewPager.getOffscreenPageLimit();
        int to = vpMyViewPager.getCurrentItem() + vpMyViewPager.getOffscreenPageLimit();
        vpMyViewPager.removeAllViews();
        for (int i = from; i <= to; i++) {
            if (i < 0) {
                continue;
            }
            adapterViewPager.instantiateItem(vpMyViewPager, i);
        }
    }
}

解释:

If you haven't changed offscreenPageLimit of the ViewPager, it always has 3 to 4 children depending on which direction you are going. And in order to show the correct content inside its children, it uses the adapter to get the right content. now when you call removeAllViews() on your ViewPager, only 3 to 4 Views are actually being removed from the Window's hierarchy, and by calling instantiateItem(ViewGroup viewPager, int index), you are only recreating 3 to 4 Views. Then everything is back to normal, you swipe and scroll, the ViewPager shows contents using its adapter. The number of its children is not the same on all devices in all situations, for example if you set the offscreenPageLimit to 5, it will probably have around 11 to 12 children, but that's all, it's not much. it's fast.

重写getItemPosition()的正确方法:

@Override
public int getItemPosition(@NonNull Object object) {
    if (!(object instanceof MyPageView)) {
        // Should never happen
        return super.getItemPosition(object);
    }
    MyDataObject dataObject = ((MyPageView) object).getData();
    if (!(dataObject instanceof MyDataObject)) {
        // Should never happen
        return super.getItemPosition(object);
    }
    if (lstItems.contains(dataObject)) {
        return POSITION_UNCHANGED;
    }
    return POSITION_NONE;
}

简而言之，您需要将用于绘制视图的数据附加/保存到视图。从那里，当调用getItemPosition()时，ViewPager可以确定当前显示的视图是否仍然存在于适配器上，并据此采取行动。