Android 如何使用ListView实现自定义BaseAdapter的筛选器？

这是从BaseAdapter继承的自定义适配器：

public class LocationItemAdapter extends BaseAdapter implements Filterable {
    private Activity context;
    private String[] names;
    private Bitmap[] iconBitmaps;
    private String[] categories;
    private String[] ratings;
    private boolean notifyChanged = true;

    public LocationItemAdapter(Activity activityContext, String[] names, Bitmap[] iconBitmaps, String[] categories, String[] ratings) {
        super();
        this.context = activityContext;
        this.names = names;
        this.iconBitmaps = iconBitmaps;
        this.categories = categories;
        this.ratings = ratings;
    }

    public int getCount() {
        return names.length;
    }

    public Object getItem(int position) {
        return position;
    }

    public long getItemId(int position) {
        return position;
    }

    public static class ItemViewHolder {
        TextView nameTextView;
        TextView categoryTextView;
        TextView ratingTextView;
        ImageView mapIconImageView;
    }

    public View getView(int position, View convertView, ViewGroup parent) {
        ItemViewHolder holder;
        LayoutInflater inflater = context.getLayoutInflater();
        if (convertView == null) {
            convertView = inflater.inflate(R.layout.location_item, null);
            holder = new ItemViewHolder();
            holder.nameTextView = (TextView) convertView.findViewById(R.id.location_item_xml_textview_name);
            holder.categoryTextView = (TextView) convertView.findViewById(R.id.location_item_xml_textview_category);
            holder.ratingTextView = (TextView) convertView.findViewById(R.id.location_item_xml_textview_rating);
            holder.mapIconImageView = (ImageView) convertView.findViewById(R.id.location_item_xml_imageview_location_icon);

            convertView.setTag(holder);
        }
        else {
            holder = (ItemViewHolder) convertView.getTag();
        }

        holder.nameTextView.setText(names[position]);
        holder.categoryTextView.setText(categories[position]);
        holder.ratingTextView.setText(ratings[position]);
        holder.mapIconImageView.setImageBitmap(iconBitmaps[position]);

        return convertView;
    }

    public void notifyDataSetChanged() {
        super.notifyDataSetChanged();
        notifyChanged = true;
    }

    public android.widget.Filter getFilter() {
        return new android.widget.Filter() {
            @SuppressWarnings("unchecked")
            @Override
            protected void publishResults(CharSequence constraint, FilterResults results) {
            }

            @Override
            protected FilterResults performFiltering(CharSequence constraint) {
                FilterResults results = new FilterResults();
                if (constraint != null && constraint.toString() > 0) {

                }
                else {
                }
                results.values = filteredResults;
                return results;
            }
        };
    }
}

我搜索发现我需要实现的两个功能是：

受保护的void publishResults（CharSequence约束、FilterResults结果）

protectedfilterResults性能过滤（字符序列约束）

---

我的筛选约束是“name”字符串数组，我猜

performFiltering

将根据文本“constraint”执行搜索。让我困惑的是

results.value

的类型是

object

。因此，在我对

名称

执行搜索之后，我应该为

结果.值

分配什么类型的数据？函数

发布结果
的作用是什么？有人能给我举一个关于这两个函数的例子吗？提前感谢，
您可以在
性能过滤中执行以下操作

        @Override
        protected void publishResults(CharSequence constraint, FilterResults results) {

              names = results.values();  // copy the results to your original names and refresh list
              notifyDataSetChanged();
        }

        @Override
        protected FilterResults performFiltering(CharSequence constraint) {
            FilterResults results = new FilterResults();
            String [] resultNames=new String[names.length];
            int count=0;

            if (constraint != null && constraint.toString() > 0) {
                  for(String name:names)
                  {
                       if(name.contains(constraint))  // whatever search condition you need
                           resultNames[count++]=name;
                  }
            }
            else {

            }
            results.values = resultNames;
            return results;
        }

你验证了下面的答案了吗。它对我非常有效。