我已经构建了DropDownmenuItem的代码,现在当我从DropDownmenuItem中单击一个项目时,它应该移动到另一个屏幕。下面是代码
class TimesScreen extends StatefulWidget {
@override
_TimesScreenState createState() => _TimesScreenState();
}
class _TimesScreenState extends State<TimesScreen> {
var gender;
@override
Widget build(BuildContext context) {
DropdownButton(
hint: Text("Select",
style: TextStyle(color: Colors.white),),
onChanged: (val){
setState(() {
this.gender=val;
});
},
value: this.gender,
items: [
DropdownMenuItem(
//onTap:
value: 'Earth',
child: Text('Earth'
),
),
DropdownMenuItem(
//onTap:
value: 'Mars',
child: Text('Mars'
),
),)]
最佳答案
您可以用具有可用于执行所需代码的Text
函数的GestureDetector
来包装您的onTap
小部件。有关更多详细信息,请查看以下内容:https://api.flutter.dev/flutter/widgets/GestureDetector-class.html
这应该有效:
DropdownMenuItem(
value: 'Earth',
child: GestureDetector(
onTap: () {
// navigate code...
},
child: Text('Earth')
),
),