Return 来自 json 的多个对象与飞镖

Question

在我正在编写的 Flutter 应用程序中，我是初学者，我想 return 来自 http 请求的 3 个对象。 http 请求工作正常，returns 数据如下：

{"one":{"fname":"Wyn","lname":"Owen","updated":1648076276673,"uid":"contact's uid","email":"wyn@wyn.com"},
"two":{"updated":1648076276673,"uid":"contact's uid","fname":"Roli","email":"roli@roli.com","lname":"Spider"},
"three":{"lname":"Poodle","email":"bill@bill.com","updated":1648076276673,"fname":"Bill","uid":"contact's uid"},
"user":{"name":null,"premium":false,"reg_id":"123456","created":1648076276673,"ads":true},
"channels":{"whatsapp":false,"email":true,"sms":false,"video":false,"app_msg":true},"shorts":{"expire":null,"code":"short_code"}}

它returns 5 KV 对。密钥将始终保持不变。我只对 3 个 KV 感兴趣：一、二和三。我想为每个 KV 对创建一个对象，并将它们称为一、二和三。我创建了以下模型：对象联系人包含 3 个人员对象：

 class Contacts {
      Contacts({
        required this.one,
        required this.two,
        required this.three,
      });
      Person one;
      Person two;
      Person three;
    
      factory Contacts.fromJson(Map<String, dynamic> json) => Contacts(
            one: json["one"],
            two: json["two"],
            three: json["three"],
          );
    
      Map<String, dynamic> toJson() => {
            "one": one.toJson(),
            "two": two.toJson(),
            "three": three.toJson(),
          };
    }
    
    class Person {
      Person({
        required this.uid,
        required this.updated,
        required this.email,
        required this.lname,
        required this.fname,
      });
    
      String uid;
      int updated;
      String email;
      String lname;
      String fname;
    
      factory Person.fromJson(Map<String, dynamic> json) => Person(
            uid: json["uid"],
            updated: json["updated"],
            email: json["email"],
            lname: json["lname"],
            fname: json["fname"],
          );
    
      Map<String, dynamic> toJson() => {
            "uid": uid,
            "updated": updated,
            "email": email,
            "lname": lname,
            "fname": fname,
          };
    }

这是class我写的：

class ContactsService {
  Future<List<Person>> fetchPersons(String uid) async {
    http.Response response =
        await http.get(Uri.parse("$PersonURL?uid=$uid"));

    if (response.statusCode == 200) {
      Map ContactData = jsonDecode(response.body);
      Person one = Person.fromJson(ContactData["one"]);
      Person two = Person.fromJson(ContactData["two"]);
      Person three = Person.fromJson(ContactData["three"]);

      List Persons = [];
      Persons.add(one);
      Persons.add(two);
      Persons.add(three);
      
      return Persons;
    } else {
      throw Exception("Something has gone wrong, ${response.statusCode}");
    }
  }
}

为了在我的脚手架中获取对象，我会放置

future: ContactsService.fetchPersons()

我希望能够引用 one.fname、two.email 等我想我在 fetchPersons() 中遗漏了一些东西。那么缺少什么？

TIA

Answer 1

你可以通过改变

来解决

Map ContactData = jsonDecode(response.body);

至

final contacts = Contacts.fromJson(jsonDecode(a));

和

Person one = Person.fromJson(ContactData["one"]);
Person two = Person.fromJson(ContactData["two"]);
Person three = Person.fromJson(ContactData["three"]);

至

final one = contacts.one;
final two = contacts.two;
final three = contacts.three;

作为补充，如果您使用 Google 提供的 JsonSerializable，则不需要序列化代码。

Return 来自 json 的多个对象与飞镖

Return multiple Objects from json with dart

json

dart

json-deserialization

flutter