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。

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

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。

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);

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!