Flutter：无法使用dio获取数组json到对象列表

Question

我试图加载数组 json（没有键的数组）并获取到对象列表中，这是我的示例代码：

我的 ApiCliet ：

Future<List<OnBoardingModel>> fetchOnboarding() async {
    try {
      Response response = await dio.get("api/OnboardingItem");
      return OnBoardingModel.fromJson(response.data) as List;
    } catch (error, stacktrace) {
      throw Exception("Exception occured: $error stackTrace: $stacktrace");
    }
  }

我的对象（使用quicktype生成）：

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

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

class OnBoardingModel {
  OnBoardingModel({
    this.id,
    this.judul,
    this.deskripsi,
    this.urlGambar,
  });

  int id;
  String judul;
  String deskripsi;
  String urlGambar;

  factory OnBoardingModel.fromJson(Map<String, dynamic> json) => OnBoardingModel(
    id: json["id"],
    judul: json["judul"],
    deskripsi: json["deskripsi"],
    urlGambar: json["urlGambar"],
  );

  Map<String, dynamic> toJson() => {
    "id": id,
    "judul": judul,
    "deskripsi": deskripsi,
    "urlGambar": urlGambar,
  };
}

最后我得到了一个错误显示：Exception: Exception occured: type 'List<dynamic>' is not a subtype of type 'Map<String, dynamic>' stackTrace: #0，参考这一行return OnBoardingModel.fromJson(response.data) as List;

你能帮我解决这个案子吗？我谢谢你

更新

我的json格式：

[
  {
    "id": 0,
    "judul": "string",
    "deskripsi": "string",
    "urlGambar": "string"
  }
]

Answer 1

最后我得到了答案，只需使用此替换返回：

return (response.data as List)
          .map((x) => OnBoardingModel.fromJson(x))
          .toList();

完整代码：

Future<List<OnBoardingModel>> fetchOnboarding() async {
    try {
      Response response = await dio.get("api/OnboardingItem");
      // if there is a key before array, use this : return (response.data['data'] as List).map((child)=> Children.fromJson(child)).toList();
      return (response.data as List)
          .map((x) => OnBoardingModel.fromJson(x))
          .toList();
    } catch (error, stacktrace) {
      throw Exception("Exception occured: $error stackTrace: $stacktrace");
    }
  }

希望这个答案可以帮助其他新来的客人。