我綁定ListView
從JSONArray
。它有一個字符串距離。我只是希望具有最小距離的列表項目必須作爲ListView
中的第一項出現,並且在最後列表項目中具有最大距離。這是我的適配器。ListView排序在Android中
public class NearByDockITAdapter extends BaseAdapter {
private Activity activity;
private ArrayList<HashMap<String, String>> data;
private static LayoutInflater inflater = null;
int loader;
public NearByDockITAdapter(Activity a, ArrayList<HashMap<String, String>> d) {
activity = a;
data = d;
inflater = (LayoutInflater) activity
.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
}
public int getCount() {
return data.size();
}
public Object getItem(int position) {
return position;
}
public long getItemId(int position) {
return position;
}
@SuppressLint("DefaultLocale")
public View getView(int position, View convertView, ViewGroup parent) {
View vi = convertView;
if (convertView == null)
vi = inflater.inflate(R.layout.list_view_item_layout, null);
TextView addressTextView = (TextView) vi
.findViewById(R.id.dockitaddresstext); // Address
HashMap<String, String> events = new HashMap<String, String>();
events = data.get(position);
String distanceString = events.get(Utility.DISTANCE);
String dis = distanceString;
addressTextView.setText("(" + dis + ") " + addressString);
return vi;
}
}
由於提前
Long.valueOf(String s).longValue();到一個列表,然後對其進行排序? – mach