flatter:List中的Json解析不是“Map类型”的子类型

Question

首先，我对处理HTTP请求和其他东西非常陌生。

我jave在JavaSpring中创建了一个API，我有一堆健身程序分配给一个用户，但是当我试图从Flutter中的JSON中获取对象时，会发生这个错误。我使用超文本传输协议和转换镖包。

列表不是类型映射的子类型

问题在于这一行：

var userModel = Users.fromJson(jsonMap);

这是getUsers方法：

Future<Users> getUsers() async{

    var client = http.Client();

    var userModel = null;


    
    var response = await client.get(Uri.parse(Strings.users_url));

    try {
      if (response.statusCode == 200) {
        var jsonString = response.body;
        var jsonMap = json.decode(jsonString);
        print(jsonMap);
        var userModel = Users.fromJson(jsonMap);

      }
    }
    catch(Exception) {
      print(Exception);
      return userModel;
    }


    return userModel;
  }

以下是用户和健身计划课程：

// To parse this JSON data, do
//
//     final users = usersFromJson(jsonString);

import 'dart:convert';

List<Users> usersFromJson(String str) => List<Users>.from(json.decode(str).map((x) => Users.fromJson(x)));

String usersToJson(List<Users> data) => json.encode(List<dynamic>.from(data.map((x) => x.toJson())));

class Users{
  Users({
    this.id,
    this.email,
    this.password,
    this.firstName,
    this.lastName,
    this.birthday,
    this.weight,
    this.height,
    this.fitnessProgrammes,
  });

  int id;
  String email;
  String password;
  String firstName;
  String lastName;
  dynamic birthday;
  int weight;
  int height;
  List<FitnessProgramme> fitnessProgrammes;

  factory Users.fromJson(Map<String, dynamic> json) => Users(
    id: json["id"],
    email: json["email"],
    password: json["password"],
    firstName: json["firstName"],
    lastName: json["lastName"],
    birthday: json["birthday"],
    weight: json["weight"],
    height: json["height"],
    fitnessProgrammes: List<FitnessProgramme>.from(json["fitnessProgrammes"].map((x) => FitnessProgramme.fromJson(x))),
  );

  Map<String, dynamic> toJson() => {
    "id": id,
    "email": email,
    "password": password,
    "firstName": firstName,
    "lastName": lastName,
    "birthday": birthday,
    "weight": weight,
    "height": height,
    "fitnessProgrammes": List<dynamic>.from(fitnessProgrammes.map((x) => x.toJson())),
  };
}

class FitnessProgramme {
  FitnessProgramme({
    this.id,
    this.name,
    this.trainingDifficulty,
  });

  int id;
  String name;
  String trainingDifficulty;

  factory FitnessProgramme.fromJson(Map<String, dynamic> json) => FitnessProgramme(
    id: json["id"],
    name: json["name"],
    trainingDifficulty: json["trainingDifficulty"],
  );

  Map<String, dynamic> toJson() => {
    "id": id,
    "name": name,
    "trainingDifficulty": trainingDifficulty,
  };
}

Answer 1

您的API可能返回一个用户数组，而不是单个用户。因此，解码的JSON类型是List

下面的代码可能会工作并实例化API返回的第一个用户：

var userModel = Users.fromJson(jsonMap[0]);