Flutter继承类如何正确生成Json_Serializable序列化代码

我定义了两个基类Student和Room，二者分别对应派生的模型类StudentModel和RoomModel。在模型类上使用json_serializable库生成JSON转换逻辑时，执行build_runner生成代码报错，相关代码和报错信息如下。

相关代码 #

import 'package:equatable/equatable.dart';
import 'package:json_annotation/json_annotation.dart';

part 'student.g.dart';

/// * Student & StudentModel

class Student extends Equatable {
  final int id;
  final String name;
  final Room studentRoom;

  const Student({this.id = 0, this.name = "", this.studentRoom = const Room()});

  @override
  List<Object?> get props => [id, name];
}

@JsonSerializable(explicitToJson: true)
class StudentModel extends Student {
  const StudentModel({int id = 0, String name = "", RoomModel studentRoom = const RoomModel()});

  factory StudentModel.fromJson(Map<String, dynamic> json) => _$StudentModelFromJson(json);

  Map<String, dynamic> toJson() => _$StudentModelToJson(this);
}

/// * Room & RoomModel

class Room extends Equatable {
  final int id;
  final String roomName;

  const Room({this.id = 0, this.roomName = ""});

  @override
  List<Object?> get props => [id, roomName];
}

@JsonSerializable()
class RoomModel extends Room {
  const RoomModel({int id = 0, String roomName = ""}) : super(id: id, roomName: roomName);

  factory RoomModel.fromJson(Map<String, dynamic> json) => _$RoomModelFromJson(json);

  Map<String, dynamic> toJson() => _$RoomModelToJson(this);
}

build_runner报错信息 #

Could not generate `toJson` code for `studentRoom`.
To support the type `Room` you can:
* Use `JsonConverter`
* Use `JsonKey` fields `fromJson` and `toJson`
package:untitled1/model/student.dart:11:14
   ╷
11 │   final Room studentRoom;
   │              ^^^^^^^^^^^
   ╵

json_serializable生成代码时是按照字段的声明类型查找序列化方法的，不会识别构造函数里传入的实际子类类型。父类Student中studentRoom字段声明为Room类型，而Room类本身没有添加@JsonSerializable注解、也没有对应的序列化逻辑，构建器找不到Room类型的fromJson/toJson方法就会抛出错误。直接在子类上加@JsonKey配置无法覆盖父类已经声明的字段类型，所以之前的尝试没有生效。

以下3种方案均可解决问题，按需选择即可：

• 方案1（最推荐）：使用泛型重构基类，类型安全无冗余
  把基类中关联的Room类型改为泛型参数，让子类在继承时明确指定关联类型为RoomModel，构建器就能直接识别到正确的可序列化类型。
  修改后的核心代码示例：

  // 基类改为泛型
  class Student<T extends Room> extends Equatable {
    final int id;
    final String name;
    final T studentRoom;
  
    const Student({this.id = 0, this.name = "", required this.studentRoom});
  
    @override
    List<Object?> get props => [id, name];
  }
  
  // 子类继承时明确指定泛型为RoomModel
  @JsonSerializable(explicitToJson: true)
  class StudentModel extends Student<RoomModel> {
    const StudentModel({
      int id = 0, 
      String name = "", 
      RoomModel studentRoom = const RoomModel()
    }) : super(id: id, name: name, studentRoom: studentRoom);
  
    factory StudentModel.fromJson(Map<String, dynamic> json) => _$StudentModelFromJson(json);
    Map<String, dynamic> toJson() => _$StudentModelToJson(this);
  }
  

  改完重新执行flutter pub run build_runner build --delete-conflicting-outputs即可正常生成代码。

• 方案2：自定义JsonConverter做类型转换
  如果不想改动原有基类结构，可以写一个专门的转换器处理Room类型和JSON的转换，逻辑里手动判断实际类型调用RoomModel的序列化方法。
  代码示例：

  class RoomConverter extends JsonConverter<Room, Map<String, dynamic>> {
    const RoomConverter();
  
    @override
    Room fromJson(Map<String, dynamic> json) {
      return RoomModel.fromJson(json);
    }
  
    @override
    Map<String, dynamic> toJson(Room object) {
      if (object is RoomModel) {
        return object.toJson();
      }
      // 兜底处理普通Room实例的序列化
      return {
        "id": object.id,
        "roomName": object.roomName,
      };
    }
  }
  

  之后把转换器加到StudentModel类上即可：

  @JsonSerializable(explicitToJson: true)
  @RoomConverter()
  class StudentModel extends Student {
    // 其余代码保持不变
  }
  

• 方案3：在子类中覆写studentRoom字段显式指定类型
  在子类中重新声明studentRoom字段，明确标注类型为RoomModel，构建器会优先识别子类覆写的字段生成序列化逻辑，注意构造函数要正确给覆写的字段赋值。
  代码示例：

  @JsonSerializable(explicitToJson: true)
  class StudentModel extends Student {
    @override
    final RoomModel studentRoom;
  
    const StudentModel({int id = 0, String name = "", this.studentRoom = const RoomModel()})
        : super(id: id, name: name, studentRoom: studentRoom);
  
    factory StudentModel.fromJson(Map<String, dynamic> json) => _$StudentModelFromJson(json);
    Map<String, dynamic> toJson() => _$StudentModelToJson(this);
  }
  

内容的提问来源于stack exchange，提问作者Sapatekno