Dart Flutter Generic Api 响应类动态类数据类型

Question

我的应用程序目前正在为每个 api 响应使用自定义类作为模型。但是我正在尝试更改它，以优化一些小东西，因此我正在尝试实现一个类包装器，例如称为 ApiResponse。但是对于 make fromJson 和 toJson，它的静态调用和方法不能正常工作。

例如，我将展示我正在尝试的内容。MyModel -> 类响应。ApiResponse -> 包含任何模型类的主类，并且必须调用子方法作为其本身“fromjson/tojson”。测试 -> 用于测试目的的类，对类的错误注释。

class MyModel {
  String id;
  String title;
  MyModel({this.id, this.title});

  factory MyModel.fromJson(Map<String, dynamic> json) {
    return MyModel(
      id: json["id"],
      title: json["title"],
    );
  }

  Map<String, dynamic> toJson() => {
        "id": this.id,
        "title": this.title,
      };
}

class ApiResponse<T> {
  bool status;
  String message;
  T data;
  ApiResponse({this.status, this.message, this.data});

  factory ApiResponse.fromJson(Map<String, dynamic> json) {
    return ApiResponse<T>(
        status: json["status"],
        message: json["message"],
        data: (T).fromJson(json["data"])); // The method 'fromJson' isn't defined for the type 'Type'.
                                           // Try correcting the name to the name of an existing method, or defining a method named 'fromJson'.
  }

  Map<String, dynamic> toJson() => {
        "status": this.status,
        "message": this.message,
        "data": this.data.toJson(), // The method 'toJson' isn't defined for the type 'Object'.
                                    // Try correcting the name to the name of an existing method, or defining a method named 'toJson'
      };
}

class Test {
  test() {
    ApiResponse apiResponse = ApiResponse<MyModel>();
    var json = apiResponse.toJson();
    var response = ApiResponse<MyModel>.fromJson(json);
  }
}

Answer 1

基本响应.dart

class BaseResponse {
  dynamic message;
  bool success;


  BaseResponse(
      {this.message, this.success});

  factory BaseResponse.fromJson(Map<String, dynamic> json) {
    return BaseResponse(
        success: json["success"],
        message: json["message"]);
  }
}

list_response.dart

server response for list
{
  "data": []
  "message": null,
  "success": true,
}

@JsonSerializable(genericArgumentFactories: true)
class ListResponse<T> extends BaseResponse {
  List<T> data;

  ListResponse({
    String message,
    bool success,
    this.data,
  }) : super(message: message, success: success);

  factory ListResponse.fromJson(Map<String, dynamic> json, Function(Map<String, dynamic>) create) {
    var data = List<T>();
    json['data'].forEach((v) {
      data.add(create(v));
    });

    return ListResponse<T>(
        success: json["success"],
        message: json["message"],
        data: data);
  }
}

single_response.dart

server response for single object
{
  "data": {}
  "message": null,
  "success": true,
}


@JsonSerializable(genericArgumentFactories: true)
class SingleResponse<T> extends BaseResponse {
  T data;

  SingleResponse({
    String message,
    bool success,
    this.data,
  }) : super(message: message, success: success);

  factory SingleResponse.fromJson(Map<String, dynamic> json, Function(Map<String, dynamic>) create) {
    return SingleResponse<T>(
        success: json["success"],
        message: json["message"],
        data: create(json["data"]));
  }
}

数据响应.dart

class DataResponse<T> {
  Status status;
  T res; //dynamic
  String loadingMessage;
  GeneralError error;

  DataResponse.init() : status = Status.Init;

  DataResponse.loading({this.loadingMessage}) : status = Status.Loading;

  DataResponse.success(this.res) : status = Status.Success;

  DataResponse.error(this.error) : status = Status.Error;


  @override
  String toString() {
    return "Status : $status \n Message : $loadingMessage \n Data : $res";
  }
}

enum Status {
  Init,
  Loading,
  Success,
  Error,
}

或者如果使用freezed那么 data_response 可以是

@freezed
abstract class DataResponse<T> with _$DataResponse<T> {
  const factory DataResponse.init() = Init;
  const factory DataResponse.loading(loadingMessage) = Loading;
  const factory DataResponse.success(T res) = Success<T>;
  const factory DataResponse.error(GeneralError error) = Error;
}

用法：（改造库的一部分

@GET(yourEndPoint)
Future<SingleResponse<User>> getUser();

@GET(yourEndPoint)
Future<ListResponse<User>> getUserList();

如果不使用 reftofit：

const _extra = <String, dynamic>{};
final queryParameters = <String, dynamic>{};
final _data = <String, dynamic>{};
final _result = await _dio.request<Map<String, dynamic>>('$commentID',
    queryParameters: queryParameters,
    options: RequestOptions(
        method: 'GET',
        headers: <String, dynamic>{},
        extra: _extra,
        baseUrl: baseUrl),
    data: _data);
final value = SingleResponse<Comment>.fromJson(
  _result.data,
  (json) => Comment.fromJson(json),
);

Answer 2

你不能在 Dart 上调用类型的方法，因为静态方法必须在编译时解析，并且类型直到运行时才有值。

但是，您可以将解析器回调传递给构造函数并使用Serializable每个模型都可以实现的接口（例如）。然后，通过更新你ApiResponse，ApiResponse<T extends Serializable>它会知道每个类型T都有一个toJson()方法。

这是更新的完整示例。

class MyModel implements Serializable {
  String id;
  String title;
  MyModel({this.id, this.title});

  factory MyModel.fromJson(Map<String, dynamic> json) {
    return MyModel(
      id: json["id"],
      title: json["title"],
    );
  }

  @override
  Map<String, dynamic> toJson() => {
        "id": this.id,
        "title": this.title,
      };
}

class ApiResponse<T extends Serializable> {
  bool status;
  String message;
  T data;
  ApiResponse({this.status, this.message, this.data});

  factory ApiResponse.fromJson(Map<String, dynamic> json, Function(Map<String, dynamic>) create) {
      return ApiResponse<T>(
      status: json["status"],
      message: json["message"],
      data: create(json["data"]),
    );
  }

  Map<String, dynamic> toJson() => {
        "status": this.status,
        "message": this.message,
        "data": this.data.toJson(),
      };
}

abstract class Serializable {
  Map<String, dynamic> toJson();
}

class Test {
  test() {
    ApiResponse apiResponse = ApiResponse<MyModel>();
    var json = apiResponse.toJson();
    var response = ApiResponse<MyModel>.fromJson(json, (data) => MyModel.fromJson(data));
  }
}