If you are saving/restoring scroll position of ListView yourself you are essentially duplicating the functionality already implemented in android framework. The ListView restores fine scroll position just well on its own except one caveat: as @aaronvargas mentioned there is a bug in AbsListView that won't let to restore fine scroll position for the first list item. Nevertheless the best way to restore scroll position is not to restore it. Android framework will do it better for you. Just make sure you have met the following conditions:

确保你没有调用setSaveEnabled(false)方法，也没有为xml布局文件中的列表设置android:saveEnabled="false"属性 为ExpandableListView重写long getCombinedChildId(long groupId, long childId)方法，使其返回正长数(BaseExpandableListAdapter类中的默认实现返回负数)。下面是一些例子:

.

@Override
public long getChildId(int groupPosition, int childPosition) {
    return 0L | groupPosition << 12 | childPosition;
}

@Override
public long getCombinedChildId(long groupId, long childId) {
    return groupId << 32 | childId << 1 | 1;
}

@Override
public long getGroupId(int groupPosition) {
    return groupPosition;
}

@Override
public long getCombinedGroupId(long groupId) {
    return (groupId & 0x7FFFFFFF) << 32;
}

如果在一个片段中使用了ListView或ExpandableListView，不要在活动重新创建片段(例如在屏幕旋转后)。使用findFragmentByTag(String标签)方法获取片段。 确保ListView有一个唯一的android:id。

To avoid aforementioned caveat with first list item you can craft your adapter the way it returns special dummy zero pixels height view for the ListView at position 0. Here is the simple example project shows ListView and ExpandableListView restore their fine scroll positions whereas their scroll positions are not explicitly saved/restored. Fine scroll position is restored perfectly even for the complex scenarios with temporary switching to some other application, double screen rotation and switching back to the test application. Please note, if you are explicitly exiting the application (by pressing the Back button) the scroll position won't be saved (as well as all other Views won't save their state). https://github.com/voromto/RestoreScrollPosition/releases

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());
          }
     }
}

我的答案是Firebase和位置0是一个变通办法

Parcelable state;

DatabaseReference everybody = db.getReference("Everybody Room List");
    everybody.addValueEventListener(new ValueEventListener() {
        @Override
        public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
            state = listView.onSaveInstanceState(); // Save
            progressBar.setVisibility(View.GONE);
            arrayList.clear();
            for (DataSnapshot messageSnapshot : dataSnapshot.getChildren()) {
                Messages messagesSpacecraft = messageSnapshot.getValue(Messages.class);
                arrayList.add(messagesSpacecraft);
            }
            listView.setAdapter(convertView);
            listView.onRestoreInstanceState(state); // Restore
        }

        @Override
        public void onCancelled(@NonNull DatabaseError databaseError) {
        }
    });

和convertView

位置0 a添加一个您不使用的空白项

public class Chat_ConvertView_List_Room extends BaseAdapter {

private ArrayList<Messages> spacecrafts;
private Context context;

@SuppressLint("CommitPrefEdits")
Chat_ConvertView_List_Room(Context context, ArrayList<Messages> spacecrafts) {
    this.context = context;
    this.spacecrafts = spacecrafts;
}

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

@Override
public Object getItem(int position) {
    return spacecrafts.get(position);
}

@Override
public long getItemId(int position) {
    return position;
}

@SuppressLint({"SetTextI18n", "SimpleDateFormat"})
@Override
public View getView(final int position, View convertView, ViewGroup parent) {
    if (convertView == null) {
        convertView = LayoutInflater.from(context).inflate(R.layout.message_model_list_room, parent, false);
    }

    final Messages s = (Messages) this.getItem(position);

    if (position == 0) {
        convertView.getLayoutParams().height = 1; // 0 does not work
    } else {
        convertView.getLayoutParams().height = RelativeLayout.LayoutParams.WRAP_CONTENT;
    }

    return convertView;
}
}

我已经看到这个工作暂时不打扰用户，我希望它为您工作

If you are saving/restoring scroll position of ListView yourself you are essentially duplicating the functionality already implemented in android framework. The ListView restores fine scroll position just well on its own except one caveat: as @aaronvargas mentioned there is a bug in AbsListView that won't let to restore fine scroll position for the first list item. Nevertheless the best way to restore scroll position is not to restore it. Android framework will do it better for you. Just make sure you have met the following conditions:

确保你没有调用setSaveEnabled(false)方法，也没有为xml布局文件中的列表设置android:saveEnabled="false"属性 为ExpandableListView重写long getCombinedChildId(long groupId, long childId)方法，使其返回正长数(BaseExpandableListAdapter类中的默认实现返回负数)。下面是一些例子:

.

@Override
public long getChildId(int groupPosition, int childPosition) {
    return 0L | groupPosition << 12 | childPosition;
}

@Override
public long getCombinedChildId(long groupId, long childId) {
    return groupId << 32 | childId << 1 | 1;
}

@Override
public long getGroupId(int groupPosition) {
    return groupPosition;
}

@Override
public long getCombinedGroupId(long groupId) {
    return (groupId & 0x7FFFFFFF) << 32;
}

如果在一个片段中使用了ListView或ExpandableListView，不要在活动重新创建片段(例如在屏幕旋转后)。使用findFragmentByTag(String标签)方法获取片段。 确保ListView有一个唯一的android:id。

To avoid aforementioned caveat with first list item you can craft your adapter the way it returns special dummy zero pixels height view for the ListView at position 0. Here is the simple example project shows ListView and ExpandableListView restore their fine scroll positions whereas their scroll positions are not explicitly saved/restored. Fine scroll position is restored perfectly even for the complex scenarios with temporary switching to some other application, double screen rotation and switching back to the test application. Please note, if you are explicitly exiting the application (by pressing the Back button) the scroll position won't be saved (as well as all other Views won't save their state). https://github.com/voromto/RestoreScrollPosition/releases

难道不是简单的android:saveEnabled="true"在ListView xml声明足够吗?

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);