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());
          }
     }
}

一个非常简单的方法:

/** Save the position **/
int currentPosition = listView.getFirstVisiblePosition();

//Here u should save the currentPosition anywhere

/** Restore the previus saved position **/
listView.setSelection(savedPosition);

方法setSelection将把列表重置为所提供的项。如果不是在触摸模式，项目将实际被选中，如果在触摸模式，项目将只定位在屏幕上。

一个更复杂的方法:

listView.setOnScrollListener(this);

//Implements the interface:
@Override
public void onScroll(AbsListView view, int firstVisibleItem,
            int visibleItemCount, int totalItemCount) {
    mCurrentX = view.getScrollX();
    mCurrentY = view.getScrollY();
}

@Override
public void onScrollStateChanged(AbsListView view, int scrollState) {

}

//Save anywere the x and the y

/** Restore: **/
listView.scrollTo(savedX, savedY);

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.

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

我使用的是FirebaseListAdapter，不能让任何解决方案工作。我最后做了这个。我猜有更优雅的方式，但这是一个完整的和有效的解决方案。

在onCreate之前:

private int reset;
private int top;
private int index;

FirebaseListAdapter内部:

@Override
public void onDataChanged() {
     super.onDataChanged();

     // Only do this on first change, when starting
     // activity or coming back to it.
     if(reset == 0) {
          mListView.setSelectionFromTop(index, top);
          reset++;
     }

 }

启动时间：

@Override
protected void onStart() {
    super.onStart();
    if(adapter != null) {
        adapter.startListening();
        index = 0;
        top = 0;
        // Get position from SharedPrefs
        SharedPreferences sharedPref = PreferenceManager.getDefaultSharedPreferences(this);
        top = sharedPref.getInt("TOP_POSITION", 0);
        index = sharedPref.getInt("INDEX_POSITION", 0);
        // Set reset to 0 to allow change to last position
        reset = 0;
    }
}

停止：

@Override
protected void onStop() {
    super.onStop();
    if(adapter != null) {
        adapter.stopListening();
        // Set position
        index = mListView.getFirstVisiblePosition();
        View v = mListView.getChildAt(0);
        top = (v == null) ? 0 : (v.getTop() - mListView.getPaddingTop());
        // Save position to SharedPrefs
        SharedPreferences sharedPref = PreferenceManager.getDefaultSharedPreferences(this);
        sharedPref.edit().putInt("TOP_POSITION" + "", top).apply();
        sharedPref.edit().putInt("INDEX_POSITION" + "", index).apply();
    }
}

因为我还必须解决这个FirebaseRecyclerAdapter，我在这里发布的解决方案:

在onCreate之前:

private int reset;
private int top;
private int index;

FirebaseRecyclerAdapter内部:

@Override
public void onDataChanged() {
    // Only do this on first change, when starting
    // activity or coming back to it.
    if(reset == 0) {
        linearLayoutManager.scrollToPositionWithOffset(index, top);
        reset++;
    }
}

启动时间：

@Override
protected void onStart() {
    super.onStart();
    if(adapter != null) {
        adapter.startListening();
        index = 0;
        top = 0;
        // Get position from SharedPrefs
        SharedPreferences sharedPref = PreferenceManager.getDefaultSharedPreferences(this);
        top = sharedPref.getInt("TOP_POSITION", 0);
        index = sharedPref.getInt("INDEX_POSITION", 0);
        // Set reset to 0 to allow change to last position
        reset = 0;
    }
}

停止：

@Override
protected void onStop() {
    super.onStop();
    if(adapter != null) {
        adapter.stopListening();
        // Set position
        index = linearLayoutManager.findFirstVisibleItemPosition();
        View v = linearLayoutManager.getChildAt(0);
        top = (v == null) ? 0 : (v.getTop() - linearLayoutManager.getPaddingTop());
        // Save position to SharedPrefs
        SharedPreferences sharedPref = PreferenceManager.getDefaultSharedPreferences(this);
        sharedPref.edit().putInt("TOP_POSITION" + "", top).apply();
        sharedPref.edit().putInt("INDEX_POSITION" + "", index).apply();
    }
}

警告! !在AbsListView中有一个错误，如果ListView.getFirstVisiblePosition()为0，则不允许onSaveState()正确工作。

所以，如果你有大图像，占据了屏幕的大部分，你滚动到第二张图像，但第一张图像的一部分正在显示，滚动位置将不会被保存…

从AbsListView.java:1650(评论我)

// this will be false when the firstPosition IS 0
if (haveChildren && mFirstPosition > 0) {
    ...
} else {
    ss.viewTop = 0;
    ss.firstId = INVALID_POSITION;
    ss.position = 0;
}

但在这种情况下，下面代码中的“top”将是一个负数，这将导致其他问题，阻止状态被正确恢复。所以当'top'为负时，就得到下一个子结点

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

if (top < 0 && getChildAt(1) != null) {
    index++;
    v = getChildAt(1);
    top = v.getTop();
}
// parcel the index and top

// when restoring, unparcel index and top
listView.setSelectionFromTop(index, top);