我使用String [] {“item 1”,“item 2”,“item 3”}作为将在我的ListView中显示的arrayadapter,但现在的问题是I不能过滤搜索目的。大多数在线教程都被过滤并从数据库中获取,但对于我的情况,我根本不触及数据库,所有这些都只是硬编码。对不起,我的英语和帮助,请:)如下如何使用searchManager搜索和过滤listview
代码: MainActivity.java:
String[] values = new String[] { "item 1", "item 2", "item 3" } ;
ItemAdapter adapter = new ItemAdapter(this, values);
setListAdapter(adapter);
Adapter.java代码如下
public class BuildingAdapter extends ArrayAdapter<String> {
private final Context context;
private final String[] buildingname;
public BuildingAdapter(Context context, String[] buildingname) {
super(context, R.layout.row_buidling_item, buildingname);
this.context = context;
this.buildingname = buildingname;
}
@Override
public View getView(int position, View convertView, ViewGroup parent) {
LayoutInflater inflater = (LayoutInflater) context
.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
View rowView = inflater.inflate(R.layout.row_buidling_item, parent, false);
TextView textView = (TextView) rowView.findViewById(R.id.buildingname);
ImageView imageView = (ImageView) rowView.findViewById(R.id.buildingicon);
TextView hints = (TextView) rowView.findViewById(R.id.buildinghints);
textView.setText(buildingname[position]);
// Change the icon for Windows and iPhone
String s = buildingname[position];
return rowView;
}
//that's all I did, hopefully someone can answer please...
如果我正确地低估了,你有2个Array,Array1和Array2,你想获得列表Array1 - Array2? –
@RohitJain先生,我只有一个数组是“buildingname”。 – MellisaNg