我正在使用兼容性库中的ViewPager。我已经成功地让它显示几个视图,我可以通过页面。

但是,我很难弄清楚如何用一组新的视图更新ViewPager。

我已经尝试了各种各样的事情,比如调用mAdapter.notifyDataSetChanged(), mviewpage .invalidate(),甚至在每次我想使用新的数据列表时创建一个全新的适配器。

没有任何帮助,textviews保持不变,从原始数据。

更新: 我做了一个小测试项目,我几乎能够更新视图。我将在下面粘贴这个类。

然而,似乎没有更新的是第二个视图,“B”仍然存在,它应该在按下更新按钮后显示“Y”。

public class ViewPagerBugActivity extends Activity {

    private ViewPager myViewPager;
    private List<String> data;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

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

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

        Button updateButton = (Button) findViewById(R.id.update_button);
        updateButton.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View v) {
                updateViewPager();
            }
        });
    }

    private void updateViewPager() {
        data.clear();
        data.add("X");
        data.add("Y");
        data.add("Z");
        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 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) {
        }
    }
}

当前回答

无论如何,在KitKat+上,适配器. notifydatasetchanged()似乎足以导致新视图显示,前提是你已经将toffscreenpagelimit设置得足够高。我能够通过执行viewpage . setoffscreenpagelimit(2)来获得所需的行为。

其他回答

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!

我认为我已经创建了一种简单的方法来通知数据集的更改:

首先,稍微改变一下instantiateItem函数的工作方式:

    @Override
    public Object instantiateItem(final ViewGroup container, final int position) {
        final View rootView = mInflater.inflate(...,container, false);
        rootView.setTag(position);
        updateView(rootView, position);
        container.addView(rootView, LayoutParams.MATCH_PARENT, LayoutParams.MATCH_PARENT);
        mViewPager.setObjectForPosition(rootView, position);
        return rootView;
    }

对于“updateView”,用你想要填充的所有数据填充视图(setText,setBitmapImage,…)

验证destroyView是这样工作的:

    @Override
    public void destroyItem(final ViewGroup container, final int position, final Object obj) {
        final View viewToRemove = (View) obj;
        mViewPager.removeView(viewToRemove);
    }

现在,假设你需要更改数据,然后调用PagerAdapter上的下一个函数:

    public void notifyDataSetChanged(final ViewPager viewPager, final NotifyLocation fromPos,
            final NotifyLocation toPos) {
        final int offscreenPageLimit = viewPager.getOffscreenPageLimit();
        final int fromPosInt = fromPos == NotifyLocation.CENTER ? mSelectedPhotoIndex
                : fromPos == NotifyLocation.MOST_LEFT ? mSelectedPhotoIndex - offscreenPageLimit
                        : mSelectedPhotoIndex + offscreenPageLimit;
        final int toPosInt = toPos == NotifyLocation.CENTER ? mSelectedPhotoIndex
                : toPos == NotifyLocation.MOST_LEFT ? mSelectedPhotoIndex - offscreenPageLimit
                        : mSelectedPhotoIndex + offscreenPageLimit;
        if (fromPosInt <= toPosInt) {
            notifyDataSetChanged();
            for (int i = fromPosInt; i <= toPosInt; ++i) {
                final View pageView = viewPager.findViewWithTag(i);
                mPagerAdapter.updateView(pageView, i);
            }
        }
    }

public enum NotifyLocation {
    MOST_LEFT, CENTER, MOST_RIGHT
}

例如,如果你希望通知viewPager所显示的所有视图,有一些东西发生了变化,你可以调用:

notifyDataSetChanged(mViewPager,NotifyLocation.MOST_LEFT,NotifyLocation.MOST_RIGHT);

就是这样。

我实际上使用notifyDataSetChanged()在ViewPager和CirclePageIndicator和之后,我调用destroyDrawingCache()在ViewPager和它的工作。其他的解决方法对我都不起作用。

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!

对我有用的是viewpage . getadapter ().notifyDataSetChanged();

在适配器中,将更新视图的代码放在getItemPosition中,就像这样

@Override
public int getItemPosition(Object object) {

    if (object instanceof YourViewInViewPagerClass) { 
        YourViewInViewPagerClass view = (YourViewInViewPagerClass)object;
        view.setData(data);
    }

    return super.getItemPosition(object);
}

可能不是最正确的方式,但它起作用了(返回POSITION_NONE技巧导致崩溃,所以不是一个选项)