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。

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!

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。

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);
    }
}

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

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.