因此,基本上我想做的是在ListView中显示从我的数据库中获得的数据,但我只能显示我的对象设备的一个属性.
这是我的设备类
class Devices {
final int iD;
final String location;
final String type;
Devices({required this.iD, required this.location, required this.type});
factory Devices.fromJson(Map<String, dynamic> json) {
return Devices(
iD: json['id'] as int,
location: json['Ort'] as String,
type: json['Typ'] as String);
}
}
我从我的PHP脚本中获得的json如下所示
{"id":1,"Ort":"Wohnzimmer","Typ":"Sensor"}
显示我的类的代码
class DeviceList extends StatelessWidget {
List<Devices> dList;
DeviceList({super.key, required this.dList});
@override
Widget build(BuildContext context) {
return ListView.builder(
itemCount: dList.length,
itemBuilder: (context, index) {
// return Text(dList[index].location,
// style: const TextStyle(color: Colors.white, fontSize: 40));
return (ListTile(title: Text(dList[index].iD.toString())));
},
);
}
}
我想显示我的类的所有属性,而不仅仅是它的ID,但我不知道如何做到这一点.