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.

Parcelable state;

@Override
public void onPause() {    
    // Save ListView state @ onPause
    Log.d(TAG, "saving listview state");
    state = listView.onSaveInstanceState();
    super.onPause();
}
...

@Override
public void onViewCreated(final View view, Bundle savedInstanceState) {
    super.onViewCreated(view, savedInstanceState);
    // Set new items
    listView.setAdapter(adapter);
    ...
    // Restore previous state (including selected item index and scroll position)
    if(state != null) {
        Log.d(TAG, "trying to restore listview state");
        listView.onRestoreInstanceState(state);
    }
}

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。

我发现了一些有趣的事情。

我尝试了setSelection和scrolltoXY，但它根本不起作用，列表仍然在相同的位置，经过一些尝试和错误，我得到了以下代码，确实工作

final ListView list = (ListView) findViewById(R.id.list);
list.post(new Runnable() {            
    @Override
    public void run() {
        list.setSelection(0);
    }
});

如果不是发布Runnable，你尝试runOnUiThread，它也不工作(至少在一些设备上)

这是一个非常奇怪的变通方法，应该是直截了当的。

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);

我发布这篇文章是因为我很惊讶没有人提到这一点。

当用户单击返回按钮后，他将返回到列表视图，在相同的状态，因为他离开它。

这段代码将覆盖“向上”按钮的行为与后退按钮相同，所以在Listview ->细节->回到Listview(没有其他选项)的情况下，这是最简单的代码来维护滚动位置和Listview中的内容。

 public boolean onOptionsItemSelected(MenuItem item) {
     switch (item.getItemId()) {
         case android.R.id.home:
             onBackPressed();
             return(true);
     }
     return(super.onOptionsItemSelected(item)); }

注意:如果你可以从细节活动转到另一个活动，向上按钮将返回到该活动，所以你必须操作后退按钮历史，以使其工作。

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

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());
          }
     }
}

private Parcelable state;
@Override
public void onPause() {
    state = mAlbumListView.onSaveInstanceState();
    super.onPause();
}

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

    if (getAdapter() != null) {
        mAlbumListView.setAdapter(getAdapter());
        if (state != null){
            mAlbumListView.requestFocus();
            mAlbumListView.onRestoreInstanceState(state);
        }
    }
}

这就够了

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

最好的解决方案是:

// 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.post(new Runnable() {
    @Override
    public void run() {
      mList.setSelectionFromTop(index, top);
   }
});

你必须在邮件和线程中调用!

用于从实现LoaderManager的ListActivity派生的活动。LoaderCallbacks使用SimpleCursorAdapter，它不能恢复onReset()中的位置，因为活动几乎总是重新启动，并且当详细信息视图关闭时适配器被重新加载。诀窍是恢复onLoadFinished()中的位置:

在onListItemClick ():

// save the selected item position when an item was clicked
// to open the details
index = getListView().getFirstVisiblePosition();
View v = getListView().getChildAt(0);
top = (v == null) ? 0 : (v.getTop() - getListView().getPaddingTop());

在onLoadFinished ():

// restore the selected item which was saved on item click
// when details are closed and list is shown again
getListView().setSelectionFromTop(index, top);

在onBackPressed ():

// Show the top item at next start of the app
index = 0;
top = 0;

对于一些正在寻找此问题解决方案的人来说，问题的根源可能在于您设置列表视图适配器的位置。在列表视图上设置适配器后，它将重置滚动位置。只是需要考虑一下。我移动设置适配器到我的onCreateView后，我们抓取引用到列表视图，它解决了我的问题。=)

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)
}

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);
}

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

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);

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();
    }
}

为了澄清Ryan Newsom的精彩回答并针对片段进行调整通常情况下，我们想要从主ListView片段导航到细节片段然后再返回主ListView片段

    private View root;
    public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState)
        {
           if(root == null){
             root = inflater.inflate(R.layout.myfragmentid,container,false);
             InitializeView(); 
           } 
           return root; 
        }

    public void InitializeView()
    {
        ListView listView = (ListView)root.findViewById(R.id.listviewid);
        BaseAdapter adapter = CreateAdapter();//Create your adapter here
        listView.setAdpater(adapter);
        //other initialization code
    }

这里的“神奇”是，当我们从细节片段导航回ListView片段时，视图不会被重新创建，我们不设置ListView的适配器，所以一切都保持不变!