对不起,英语不好。我想根据gridview项的ID从一个屏幕导航到另一个屏幕。填写我的代码:
class CategoryScreen extends StatelessWidget {
@override
Widget build(BuildContext context) {
return new SafeArea(
child: new Scaffold(
appBar: new BankAndaAppBar(),
body: new Container(
color: Colors.grey[100],
child: new ListView(
physics: ClampingScrollPhysics(),
children: <Widget>[
new Container(
padding: EdgeInsets.only(left: 12.0, right: 12.0, top: 12.0),
color: Colors.grey[100],
child: new Column(
children: <Widget>[
_bodyGridView(),
],
)),
],
),
),
),
);
}
Widget _bodyGridView() {
return SingleChildScrollView(
child: new Column(
children: <Widget>[
GridView.count(
shrinkWrap: true,
physics: NeverScrollableScrollPhysics(),
crossAxisCount: 3,
children: CATEGORIES_DUMMY_DATA
.map((cat) => ListCategory(cat.id, cat.title, cat.titleDetail, cat.description, cat.image))
.toList(),
),
],
),
);
}
}
我想在单击网格视图项目后显示详细信息。示例:每次单击/轻击gridview项1都基于id = 1显示详细视图。