如何在Dart中实现支持泛型T的通用JSON解析类？

问题背景 #

现有两种JSON响应格式：

Response 1 #

{
   "status" :"ok",
   "message":"found",
   "data" : {
        "key1" :"value1",
        "key2" :"value2"
   }
}

Response 2 #

{
  "status": "ok",
  "message": "found",
  "data": {
    "users": [
      {
        "key1": "value1",
        "key2": "value2"
      }
    ]
  }
}

需要创建带泛型T的Common类，实现类似Java泛型的效果，支持以下两种解析方式：

Common<List<Data>> commonRes = Common<List<Data>>.fromJson(jsonDecode(res.body));

或

Common<Data> commonRes = Common<Data>.fromJson(jsonDecode(res.body));

尝试编写的Common类未达到预期效果：

class Common<T> {
  String? status;
  String? message;
  T? data;

  Common({
    this.status,
    this.message,
    this.data,
  });

  Common.fromJson(Map<String, dynamic> json) {
    status = json['status'];
    message = json['message'];
    data = json['data'] as T;
  }

  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data1 = Map<String, dynamic>();
    data1['status'] = this.status;
    data1['message'] = this.message;
    data1['data'] = this.data;

    return data1;
  }
}

问题原因与解决方法 #

原代码直接用as T强转json['data']无法处理复杂类型（如自定义对象、List<Data>），因为Dart泛型存在运行时擦除，无法自动推断如何将Map<String, dynamic>转为T类型。

核心解决思路是添加类型转换回调函数，让调用者传入json['data']到T的转换逻辑。

正确实现代码 #

class Common<T> {
  String? status;
  String? message;
  T? data;

  Common({
    this.status,
    this.message,
    this.data,
  });

  // 新增带转换函数的构造函数，由外部定义data字段的解析逻辑
  Common.fromJson(Map<String, dynamic> json, T Function(dynamic) fromJsonT) {
    status = json['status'];
    message = json['message'];
    data = fromJsonT(json['data']);
  }

  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data1 = <String, dynamic>{};
    data1['status'] = status;
    data1['message'] = message;
    data1['data'] = data;
    return data1;
  }
}

使用示例 #

假设已定义Data实体类：

class Data {
  String? key1;
  String? key2;

  Data({this.key1, this.key2});

  Data.fromJson(Map<String, dynamic> json) {
    key1 = json['key1'];
    key2 = json['key2'];
  }

  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = <String, dynamic>{};
    data['key1'] = key1;
    data['key2'] = key2;
    return data;
  }
}

1. 解析Response 1（单个Data对象）：

Common<Data> commonRes = Common<Data>.fromJson(
  jsonDecode(res.body),
  (json) => Data.fromJson(json as Map<String, dynamic>),
);

2. 解析Response 2（users列表）：

Common<List<Data>> commonRes = Common<List<Data>>.fromJson(
  jsonDecode(res.body),
  (json) {
    List<dynamic> userList = json['users'] as List;
    return userList.map((item) => Data.fromJson(item as Map<String, dynamic>)).toList();
  },
);

简化写法（可选） #

针对列表场景，可以添加辅助构造函数简化调用：

// 针对List类型的辅助构造函数
Common.fromJsonList(Map<String, dynamic> json, T Function(dynamic) fromJsonItem) {
  status = json['status'];
  message = json['message'];
  List<dynamic> list = json['data']['users'] as List;
  data = list.map((item) => fromJsonItem(item)).toList() as T;
}

使用时：

Common<List<Data>> commonRes = Common.fromJsonList(
  jsonDecode(res.body),
  (item) => Data.fromJson(item as Map<String, dynamic>),
);

内容的提问来源于stack exchange，提问作者Hardik Mehta