按字母顺序对 ListView 的元素进行排序

Question

我正在尝试使用 Comparable java 接口及其方法 Compareto() 按字母顺序对添加到 ListView 的元素进行排序。

这是我实现的适配器的代码：我尝试在 addLocation() 中使用 Collections.sort 方法。

您可以在this post查看我项目的其他文件

public class LocationAdapter extends BaseAdapter {

        private List<Location> Locations;
        int monLayout;
        LayoutInflater inflater;


        public LocationAdapter(Context context, int layout){
            monLayout=layout;
            inflater = (LayoutInflater)context.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
            Locations = new ArrayList<Location>();
        }



        private class Location implements Comparable{
            public String name;
            public String address;
            public Long date;

            public Location(String _name,String _address , Long _date){
                name=_name;
                address=_address;
                date=_date;

            }

            @Override
            public int compareTo(Object another) {

                int res;

                Location loc=(Location)another;
                char firstLetter = loc.name.charAt(0);
                char thisFirstLetter = this.name.charAt(0);

                if(thisFirstLetter<firstLetter){
                    res=1;
                }
                else if(thisFirstLetter<firstLetter){
                    res=-1;
                }
                else
                {
                    res=0;
                }

                return res;
            }
        }


        private class ViewHolder{

            TextView name_view;
            TextView address_view;
            TextView date_view;

            public ViewHolder(View rowLayout){

                name_view = (TextView)rowLayout.findViewById(R.id.name);
                date_view = (TextView)rowLayout.findViewById(R.id.date);
                address_view = (TextView)rowLayout.findViewById(R.id.address);
            }
        }

        public void addLocation(String _name,String _address,Long _date){
            //Création d'une nouvelle location avec les données en paramètres
            Location new_loc = new Location(_name, _address,_date);

            //Ajout de la location à la liste des locations
            Locations.add(new_loc);
            Collections.sort(Locations);
        }
    /*Méthodes de la classe mère*/

    @Override
    public int getCount() {
        return Locations.size();
    }

    @Override
    public Object getItem(int position) {
        return Locations.get(position);
    }

    @Override
        public long getItemId(int position) {
            return 0;
        }

        @Override
        public View getView(int position, View convertView, ViewGroup parent)
        {
            View view = convertView;

            if (view == null)
                view = inflater.inflate(monLayout, parent, false);

            ViewHolder holder = (ViewHolder)view.getTag();
            if (holder == null)
            {
                holder = new ViewHolder(view);
                view.setTag(holder);
            }

            Location location = (Location)getItem(position);

            holder.name_view.setText(location.name);
            holder.address_view.setText(location.address);


            Date date=new Date(location.date);

            SimpleDateFormat df = (SimpleDateFormat) SimpleDateFormat.getDateInstance(SimpleDateFormat.MEDIUM);
            String dateText = df.format(date);
            holder.date_view.setText(dateText);


            return view;
        }
}

Answer 1

使用自定义Comparator：

Collections.sort(list, new Comparator<String>() {
        @Override
        public int compare(String s1, String s2) {
            return s1.compareToIgnoreCase(s2);
        }
    });

或

Collections.sort(list, String.CASE_INSENSITIVE_ORDER);

在您的订单中，您可以这样做：

删除implements 并使用它来代替你的Collections.sort(Locations)；

Collections.sort(Locations, new Comparator<Location>() {
        @Override
        public int compare(final Location object1, final Location object2) {
            return object1.name.compareTo(object2.name);
        }
    } 
);