json - flutter json解析，能够部分解析

Question

以下是我的 json 输出

{
    "success": true,
    "data": {
        "id": 1,
        "username": "MV",
        "name": "Saraa",
        "userSponsorData": {
            "initial": "GT",
            "name": "Green Trust",
            "email": "trust@mailinator.com",
            "username": "trust",
            "phone": "1234567890"
        }
    }
}

以下是我的模型 Dart 文件的一部分

class UserDetailModel {
  bool success;
  Data data;

  UserDetailModel({this.success, this.data});

  UserDetailModel.fromJson(Map<String, dynamic> json) {
    success = json['success'];
    data = json['data'] != null ? new Data.fromJson(json['data']) : null;
  }
}

通过解析我能够获得jsonObject“data”下的值，这些值能够被打印

class Data {
  String name;
  String email;
  String uname;
  String phone;
  SponsorData sData;

  Data({this.name, this.email, this.uname, this.phone, this.sData});

  Data.fromJson(Map<String, dynamic> json) {
    name = json['name'];
    email = json['email'];
    uname = json['username'];
    phone = json['phone'];
    sData = json['userSponsorData'] != null
        ? new SponsorData.fromJson(json['userSponsorData'])
        : null;
  }
}

但无法从 jsonObject“userSponsorData”中检索 - 当我尝试打印值时，

NoSuchMethodError: The getter 'initial' was called on null. - how to fix this

Answer 1

void main() {
    var value = {
      "success": true,
      "data": {
        "id": 1,
        "username": "MV",
        "name": "Saraa",
        "userSponsorData": {
          "initial": "GT",
          "name": "Green Trust",
          "email": "trust@mailinator.com",
          "username": "trust",
          "phone": "1234567890"
        }
      }
    };
    new Data.fromJson(value['data']);
}

class Data {
  String name;
  String email;
  String uname;
  String phone;
  SponsorData sData;

  Data({this.name, this.email, this.uname, this.phone, this.sData});

  Data.fromJson(Map<String, dynamic> json) {
    name = json['name'];
    email = json['email'];
    uname = json['username'];
    phone = json['phone'];
    sData = json['userSponsorData'] != null
        ? new SponsorData.fromJson(json['userSponsorData'])
        : null;
  }
}

class SponsorData {
  String sdata;
  SponsorData({this.sdata});
  SponsorData.fromJson(Map<String, dynamic> json) {
    var initial = json['initial'];
    var name = json['name'];
    var email = json['email'];
    var username = json['username'];
    var phone = json['phone'];
    print("$initial , $name , $email");
  }
}

成功结果 ojita