我有一个很长的ListView,用户可以在返回前一个屏幕之前滚动它。当用户再次打开这个ListView时,我希望列表被滚动到与之前相同的位置。关于如何实现这一点,你有什么想法吗?


当前回答

试试这个:

// save index and top position
int index = mList.getFirstVisiblePosition();
View v = mList.getChildAt(0);
int top = (v == null) ? 0 : (v.getTop() - mList.getPaddingTop());

// ...

// restore index and position
mList.setSelectionFromTop(index, top);

Explanation: ListView.getFirstVisiblePosition() returns the top visible list item. But this item may be partially scrolled out of view, and if you want to restore the exact scroll position of the list you need to get this offset. So ListView.getChildAt(0) returns the View for the top list item, and then View.getTop() - mList.getPaddingTop() returns its relative offset from the top of the ListView. Then, to restore the ListView's scroll position, we call ListView.setSelectionFromTop() with the index of the item we want and an offset to position its top edge from the top of the ListView.

其他回答

这里提供的解决方案似乎都不适合我。在我的情况下,我有一个ListView在一个片段,我替换在一个FragmentTransaction,所以一个新的片段实例创建每次片段显示,这意味着ListView状态不能存储为片段的成员。

相反,我最终将状态存储在我的自定义Application类中。下面的代码应该会让你了解它是如何工作的:

public class MyApplication extends Application {
    public static HashMap<String, Parcelable> parcelableCache = new HashMap<>();


    /* ... code omitted for brevity ... */
}

 

public class MyFragment extends Fragment{
    private ListView mListView = null;
    private MyAdapter mAdapter = null;


    @Override
    public void onViewCreated(View view, @Nullable Bundle savedInstanceState) {
        super.onViewCreated(view, savedInstanceState);

        mAdapter = new MyAdapter(getActivity(), null, 0);
        mListView = ((ListView) view.findViewById(R.id.myListView));

        Parcelable listViewState = MyApplication.parcelableCache.get("my_listview_state");
        if( listViewState != null )
            mListView.onRestoreInstanceState(listViewState);
    }


    @Override
    public void onPause() {
        MyApplication.parcelableCache.put("my_listview_state", mListView.onSaveInstanceState());
        super.onPause();
    }

    /* ... code omitted for brevity ... */

}

基本思想是将状态存储在片段实例之外。如果您不喜欢在应用程序类中拥有静态字段的想法,我猜您可以通过实现一个片段接口并将状态存储在您的活动中来实现它。

另一种解决方案是将其存储在SharedPreferences中,但这有点复杂,您需要确保在应用程序启动时清除它,除非您希望在应用程序启动时保持状态。

 

另外,为了避免“当第一项可见时滚动位置不保存”,你可以显示一个0px高度的虚拟第一项。这可以通过重写适配器中的getView()来实现,如下所示:

@Override
public View getView(int position, View convertView, ViewGroup parent) {
    if( position == 0 ) {
        View zeroHeightView = new View(parent.getContext());
        zeroHeightView.setLayoutParams(new ViewGroup.LayoutParams(0, 0));
        return zeroHeightView;
    }
    else
        return super.getView(position, convertView, parent);
}

如果在重新加载前保存状态,并在重新加载后恢复状态,则可以在重新加载后保持滚动状态。在我的情况下,我做了一个异步网络请求,并在它完成后在回调中重新加载列表。这是我恢复状态的地方。代码示例是Kotlin。

val state = myList.layoutManager.onSaveInstanceState()

getNewThings() { newThings: List<Thing> ->

    myList.adapter.things = newThings
    myList.layoutManager.onRestoreInstanceState(state)
}

使用下面的代码:

int index,top;

@Override
protected void onPause() {
    super.onPause();
    index = mList.getFirstVisiblePosition();

    View v = challengeList.getChildAt(0);
    top = (v == null) ? 0 : (v.getTop() - mList.getPaddingTop());
}

无论何时你刷新你的数据使用下面的代码:

adapter.notifyDataSetChanged();
mList.setSelectionFromTop(index, top);

如果你在一个活动上使用片段,你可以这样做:

public abstract class BaseFragment extends Fragment {
     private boolean mSaveView = false;
     private SoftReference<View> mViewReference;

     @Override
     public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
          if (mSaveView) {
               if (mViewReference != null) {
                    final View savedView = mViewReference.get();
                    if (savedView != null) {
                         if (savedView.getParent() != null) {
                              ((ViewGroup) savedView.getParent()).removeView(savedView);
                              return savedView;
                         }
                    }
               }
          }

          final View view = inflater.inflate(getFragmentResource(), container, false);
          mViewReference = new SoftReference<View>(view);
          return view;
     }

     protected void setSaveView(boolean value) {
           mSaveView = value;
     }
}

public class MyFragment extends BaseFragment {
     @Override
     public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
          setSaveView(true);
          final View view = super.onCreateView(inflater, container, savedInstanceState);
          ListView placesList = (ListView) view.findViewById(R.id.places_list);
          if (placesList.getAdapter() == null) {
               placesList.setAdapter(createAdapter());
          }
     }
}

我采用了@(Kirk Woll)建议的解决方案,它对我很有效。我还在“联系人”应用程序的Android源代码中看到,他们使用了类似的技术。我还想补充一些具体情况: 在我的listactivity派生类的顶部:

private static final String LIST_STATE = "listState";
private Parcelable mListState = null;

然后,一些方法重写:

@Override
protected void onRestoreInstanceState(Bundle state) {
    super.onRestoreInstanceState(state);
    mListState = state.getParcelable(LIST_STATE);
}

@Override
protected void onResume() {
    super.onResume();
    loadData();
    if (mListState != null)
        getListView().onRestoreInstanceState(mListState);
    mListState = null;
}

@Override
protected void onSaveInstanceState(Bundle state) {
    super.onSaveInstanceState(state);
    mListState = getListView().onSaveInstanceState();
    state.putParcelable(LIST_STATE, mListState);
}

当然,“loadData”是我从DB中检索数据并将其放入列表的函数。

在我的Froyo设备上,当你改变手机方向时,当你编辑一个项目并返回列表时,这都是有效的。