使用fromJson扑空安全性 [英] Flutter null safety with fromJson

问题描述

最近要迁移到Flutter的null安全功能，我有很多需要更新的类.

Recently migrating to Flutter null safety feature, I have a lot of Classes that I need to update.

对于我的模型，我使用 fromJson 来反序列化来自json对象的数据.这迫使我为每个非必填字段输入 late 关键字.

For my models, I use fromJson to deserialize the data from a json object. This forces me to put the late keyword for each field that is non optional.

这是正确的方法吗?

class ServerSession {
  late String sessionId;
  late String refreshToken;
  late String accessToken;

  ServerSession({required this.sessionId, required this.refreshToken, required this.accessToken});

  ServerSession.fromJson(Map<String, dynamic> json) {
    sessionId = json['session_id'] ?? json['sessionId'];
    refreshToken = json['refresh_token'] ?? json['refreshToken'];
    accessToken = json['access_token'] ?? json['accessToken'];
  }

  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = new Map<String, dynamic>();
    data['sessionId'] = this.sessionId;
    data['refreshToken'] = this.refreshToken;
    data['accessToken'] = this.accessToken;
    return data;
  }
}

推荐答案

不，不是.您应该使用初始化列表来初始化类的字段.您可以在语言导览中了解有关初始化程序列表的更多信息.

No, it is not. You should be using the initializer list for initializing fields of your class. You can read more about the initializer list in the language tour.

class ServerSession {
  String sessionId;
  String refreshToken;
  String accessToken;

  ServerSession({required this.sessionId, required this.refreshToken, required this.accessToken});

  ServerSession.fromJson(Map<String, dynamic> json) :
    sessionId = json['session_id'] ?? json['sessionId'],
    refreshToken = json['refresh_token'] ?? json['refreshToken'],
    accessToken = json['access_token'] ?? json['accessToken'];

  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = new Map<String, dynamic>();
    data['sessionId'] = this.sessionId;
    data['refreshToken'] = this.refreshToken;
    data['accessToken'] = this.accessToken;
    return data;
  }
}

---

我个人会使用正常"对象的构造函数，并使 fromJson 成为工厂构造函数，尽管这两种方法都可以.

---

I personally would use the "normal" constructor for the object and make fromJson a factory constructor, though either method works.

class ServerSession {
  String sessionId;
  String refreshToken;
  String accessToken;

  ServerSession({required this.sessionId, required this.refreshToken, required this.accessToken});

  factory ServerSession.fromJson(Map<String, dynamic> json) {
    return ServerSession(
      sessionId: json['session_id'] ?? json['sessionId'],
      refreshToken: json['refresh_token'] ?? json['refreshToken'],
      accessToken: json['access_token'] ?? json['accessToken']
    );
  }

  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = new Map<String, dynamic>();
    data['sessionId'] = this.sessionId;
    data['refreshToken'] = this.refreshToken;
    data['accessToken'] = this.accessToken;
    return data;
  }
}

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