从Firestore获取Flutter CupertinoPicker值

Question

我一直在努力使用Flutter上的CupertinoPicker将值整数转换为字符串，我的代码看起来有点像：

class AddExerciseMuscleGroupPicker extends StatefulWidget {
const AddExerciseMuscleGroupPicker({Key? key}) : super(key: key);
@override
_AddExerciseMuscleGroupPickerState createState() => _AddExerciseMuscleGroupPickerState();
}
class _AddExerciseMuscleGroupPickerState extends State<AddExerciseMuscleGroupPicker> {
final Stream<QuerySnapshot> _usersStream = FirebaseFirestore.instance.collection('MuscleGroups').orderBy('MuscleGroupName').snapshots();
@override
Widget build(BuildContext context) {
return StreamBuilder<QuerySnapshot>(
stream: _usersStream,
builder: (BuildContext context, AsyncSnapshot<QuerySnapshot> snapshot) {
if (snapshot.hasError) {
return Text('Something went wrong');
}
if (snapshot.connectionState == ConnectionState.waiting) {
return Text("Loading");
}
return CupertinoPicker(
itemExtent: 30,
children: snapshot.data!.docs.map((DocumentSnapshot document) {
Map<String, dynamic> data = document.data()! as Map<String, dynamic>;
return Text(data['MuscleGroupName']);
}).toList(),
onSelectedItemChanged: (value) {
setState(() {
newExerciseMuscleGroupController.text = value.toString();
});
},
);
},
);
}
}

就目前而言，我可以创建"；肌肉群"；它们会出现在CupertinoPicker上，然而，一旦我选择了一个"；肌肉群"；，这将向TextEditingController返回一个整数，我的问题是，如何返回字符串而不是整数？如果这种方法是错误的，我很感激任何小的指导来解决这个问题。如果我得到的列表是"；胸部"Abs"腿"；，它返回"0"；0"1〃"2〃；而不是"；胸部"Abs"；或"；腿；

Answer 1

您可以设置snapshot.data！[value]到controller.text.

//您不必包装controller.text在setState中的值更改。它会更新TextField。