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!

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.

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!

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。=)

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!