方案1： 针对需要更新的item调用public View getView(int position, View convertView, ViewGroup parent)即可。 如：
public class AuthorListAdapter extends BaseAdapter {
...
@Override
public View getView(int position, View convertView, ViewGroup parent) {
...
return convertView;
}
/**
* 更新Item视图，减少不必要的重绘
*
* @param listView
* @param position
*/
public void updateItemView(ListView listView, int position) {
//换算成 Item View 在 ViewGroup 中的 index
int index = position - listView.getFirstVisiblePosition();
if (index >= 0 && index < listView.getChildCount()) {
//更新数据
AuthorInfo authorInfo = mAuthorInfoList.get(position);
authorInfo.setNickName("Google Android");
authorInfo.setMotto("My name is Android .");
authorInfo.setPortrait(R.mipmap.ic_launcher);
//更新单个Item
View itemView = listView.getChildAt(index);
getView(position, itemView, listView);
}
}
}
注意：在ListView中，getChildAt（int position)返回的item是指的可视区域内的第position个元素。使用getChildAt(index)的取值，只能是当前可见区域（列表可滚动）的子项！
即取值范围在 >= ListView.getFirstVisiblePosition() && <= ListView.getLastVisiblePosition();
这里涉及到listview的复用原理，简单来说：一个元素很多的listview，如果页面的可视区域内最多可以看到6个元素，假如它们的index是0-5,那么在内存中只有6个item对象。当第7个item(即index=6)进入到可视区域的时候，那么其实是在复用index=0的item。由于在滑动时getView这个更新控件的方法执行非常频繁，所以肉眼是丝毫看不出这种复用的。
方案2@Override
public View getView(int position, View convertView, ViewGroup parent)
{
View view = super.getView(position, convertView, parent);
view.setTag(getItemId(position));
return view;
}
For the update check every element of list, if a view with given id is there it's visible so we perform the update.
private void update(long id)
{ int c = list.getChildCount();
for (int i = 0; i < c; i++)
{
View view = list.getChildAt(i);
if ((Long)view.getTag() == id)
{
// update view
}
}
}
It's actually easier than other methods and better when you dealing with ids not positions! Also you must call update for items which get visible.