Saving a class within a class - Flutter - flutter

Question

I have class "DilemmData" which I need to save:
class DilemmData {
double percent = 0;
final String title;
final String date;
List<DilemmItemData> plus = [];
List<DilemmItemData> minus = [];
DilemmData(
plus,
minus, {
this.percent = 0,
this.title = '',
this.date = '',
});
static Map<String, dynamic> toJson(DilemmData dilemm) {
return {
'percent': dilemm.percent,
'title': dilemm.title,
'date': dilemm.date,
'plus': dilemm.plus.map((e) => e.toJson()).toList(),
'minus': dilemm.minus.map((e) => e.toJson()).toList(),
};
}
factory DilemmData.fromJson(Map<String, dynamic> json) {
return DilemmData(
json['plus'],
json['minus'],
percent: json['percent'],
title: json['title'],
date: json['date'],
);
}
static String encode(List<DilemmData> pressure) => json.encode(
pressure
.map<Map<String, dynamic>>((dil) => DilemmData.toJson(dil))
.toList(),
);
static List<DilemmData> decode(String dilemm) =>
((json.decode(dilemm) ?? []) as List<dynamic>)
.map<DilemmData>((dil) => DilemmData.fromJson(dil))
.toList();
}
class DilemmItemData {
final int importance;
final String argument;
DilemmItemData({this.importance = 0, this.argument = ''});
Map<String, dynamic> toJson() {
return {
'importance': importance,
'argument': argument,
};
}
factory DilemmItemData.fromJson(Map<String, dynamic> json) {
return DilemmItemData(
importance: json['importance'], argument: json['argument']);
}
}
There is save function:
DilemmData dilemm = DilemmData(
percent: 0,
title: controller.text,
date: clockString);
SharedPreferences sharedPreferences =
await SharedPreferences.getInstance();
String data = jsonEncode(dilemm);
sharedPreferences.setString('dilemms', data);
But when i try to save i get this error: JsonUnsupportedObjectError (Converting object to an encodable object failed: Instance of 'DilemmData'). Does anyone know how to save?
StackOverflow says I have a lot of code, but I don't know what else to write, I'll write random letters for this: shdjfhjjsdfjhwehfwouiehuwefuwefuwheugweghuweghuiweghuwueghuweweugihwueighuwhguwhgu
Here is the function to get the data:
Future loadDillems() async {
SharedPreferences sharedPreferences = await
SharedPreferences.getInstance();
if (sharedPreferences.containsKey('dilemms')) {
Map<String, dynamic> data =
jsonDecode(sharedPreferences.getString('dilemms')!);
MyApp.dilemmList = DilemmData.fromJson(data) as List<DilemmData>;
}
}

Answer 1

When converting the "top" class using its toJson() method, make sure that the subsequent classes' toJson() methods is also called. As such:
'plus': dilemm.plus.map((e) => e.toJson()).toList(),
And you of course have to implement toJson() and fromJson() in DilemmItemData as well.
Edit:
Okey, so a couple of alternatives. First of all, you had to create the new toJson() (and soon also fromJson()) methods. That is good.
As #Ivo wrote, if you pass the object to jsonEncode as you do now, then the methods cannot be static. But you could also pass DilemmData.toJson(dilemm) as:
String data = jsonEncode(DilemmData.toJson(dilemm));
But I'd recommend as #Ivo wrote to make it non-static, as you did with the new toJson method, and of course keep the toJson in DilemmItemData, and I suggest that you write the fromJson as well...
Second edit:
Your fromJson have to be a bit sharper. Something like this:
DilemmData(
(json['plus'] as List<dynamic>)
.map((e) => DilemmItemData.fromJson(e as Map<String, dynamic>))
.toList(),
Edit 3:
You are saving one DilemmData, so you have to read it as one (1) DilemmData. Remove the as List<DilemmData>; It is not all of a sudden a List when it is one object that you save (String data = jsonEncode(dilemm);)
So do as follows:
final oneDilemmData = DilemmData.fromJson(data);

Answer 2

I think you need to make your toJson() non static like this:
Map<String, dynamic> toJson() {
return {
'percent': percent,
'title': title,
'date': date,
'plus': plus,
'minus': minus,
};
}
your encode could be like this then:
static String encode(List<DilemmData> pressure) => json.encode(
pressure
.map<Map<String, dynamic>>((dil) => dil.toJson())
.toList(),
);