Dart/Flutter 如何实现一个有两个版本（PRO 和 LITE）的颤振项目

Question

In my flutter project I have some features that I only want to have on a PRO version of my app. How can I build two different final projects, one with the feature and another without?

Answer 1

in your user model, add a field like "status", and then set it to lite by default, then in your UI, after user logs in, grab the currentUserData, and instantiate the User, then check if User.status is pro or lite then build as such

@override
Widget build(BuildContext context) {
  return user.status == lite ? LiteHomeScreen() : ProHomeScreen();
}

Long Example

User

enum Status {
  lite,
  pro,
}

class User {
  String? name;
  int? age;
  Status status;
  
  User({this.name, this.age, this.status = Status.lite});

  Map<String, dynamic> toJson() {
    return {
      'name': name,
      'age': age,
      'status': status,
    };
  }

  factory User.fromJson(Map<String, dynamic> json) {
    return User(
      name: json['name'],
      age: json['age'],
      status: json['status'],
    );
  }

  @override
  bool operator ==(Object other) =>
      identical(this, other) ||
      other is User &&
          runtimeType == other.runtimeType &&
          name == other.name &&
          age == other.age &&
          status == other.status;

  @override
  int get hashCode => name.hashCode ^ age.hashCode ^ status.hashCode;
}

UI

class Home extends StatefulWidget {
  const Home({Key? key}) : super(key: key);

  @override
  State<Home> createState() => _HomeState();
}

class _HomeState extends State<Home> {
  late User user;
  void initializeUser() async {
    Response response = await get("https://fakedatabase.myfake.com/users/1");
    user = User.fromJson(jsonDecode(response.body));
  }
  @override
  void initState() {
    initializeUser();
    super.initState();
  }
  @override
  Widget build(BuildContext context) {
    return user.status == Status.lite ? LiteHome() : ProHome();
  }
}

then write a method that will execute when user pays for pro or something

void upgradeToPro() {
  user.status = Status.pro;
  
  /// Save the user to the database.
  post(...uri, body: jsonEncode(user.toJson()));
}