smoothScrollToPosition in ExpandableListView

Question

我试图用ExpandableListView方法滚动到smoothScrollToPosition(int position)的底部，但是失败了，因为我不知道如何计算位置属性。

都不是

adapter.getGroupCount()

nor

adapter.getChildrenCount(getGroupCount()-1);

nor

public int getLastIndex() {
    int count = 0;
    for (int g = 0; g < getGroupCount(); g++)
        count += getChildrenCount(g) + 1;
    return count;
}

成功了。

TLDR:我如何在API级别的ExpandableListView上滚动到8+的底部？

Answer 1

expandableListView.setOnGroupClickListener(new OnGroupClickListener() {

    @Override
    public boolean onGroupClick(ExpandableListView parent, View v,
            int groupPosition, long id) {
        // TODO Auto-generated method stub
        parent.smoothScrollToPosition(groupPosition);
        return true;
    }
});

Answer 2

listView.post(new Runnable() {
    @Override
    public void run() {
        listView.smoothScrollToPositionFromTop(listView.getCount()*2, 0, 5000);
    }
});

ExpandableListView.post()用于等待列表完成项目加载。获取项目(组和子组)的计数的ExpandableListView.getCount()

提供持续时间，这样列表就不会快速滚动(方法需要API 11) smoothScrollToPositionFromTop(int position, int offset, int duration)

Answer 3

你的职位计算如下：

public int getLastIndex() {
    int count = 0;
    for (int g = 0; g < getGroupCount(); g++)
        count += getChildrenCount(g) + 1;
    return count;
}

你可以用它滚动

expandableListView.smoothScrollToPosition(getLastIndex());

但是，如果在活动类的smoothScrollToPosition()方法中调用onCreate()方法，则应该在界面显示后使用Handler调用smoothScrollToPosition()方法：

  @Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    .....
    new Handler().postDelayed(new Runnable() {
        @Override
        public void run() {
            expandableListView.smoothScrollToPosition(getLastIndex());
        }
    },1000);
}