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我有一個GridView與ImageView項目,我想在某個位置以編程方式選擇一個項目。我認爲這很容易,但是... setSelection
什麼都不做,爲什麼?Android - 如何選擇GridView中的項目?
public class Test extends Activity {
public void onCreate (final Bundle savedInstanceState) {
super.onCreate (savedInstanceState);
GridView gridView = new GridView (this);
gridView.setNumColumns (4);
gridView.setHorizontalSpacing (16);
gridView.setVerticalSpacing (16);
setContentView (gridView);
gridView.setAdapter (new BaseAdapter() {
public int getCount() {
return 12;
}
public Object getItem (int position) {
return position;
}
public long getItemId (int position) {
return position;
}
public View getView (int position, View convertView, ViewGroup parent) {
ImageView view = (ImageView) convertView;
if (view == null) {
view = new ImageView (Test.this);
view.setImageResource (R.drawable.image);
}
return view;
}
});
gridView.setSelection (3); // ?? does nothing
}
}
也許添加gridView.setSelection(3 ); setContentView(gridView)之後; –
@SirukViktor嘗試了這一點,但它沒有改變任何東西...... – Patrick
也許嘗試在gridview中定義一個你的項目,然後給這個項目onClickListener。你是否有理由在你的xml中聲明你的javaview中的gridview? – ksudu94