如果将来添加任何键，json如何解析动态键和值

Question

我在最初获得不同形状属性的地方得到响应。

{
    "success": true,
    "data": {
         "pcode": "001",
         "pname": "ABC",
         "pprice": "100"
        "ones": [{
                "id": "2",
                "username": "LM10002"
            },
            {
                "id": "6",
                "username": "LM10006"
            }
        ],
        "twos": [{
                "id": "3",
                "username": "LM10003"
            },
            {
                "id": "8",
                "username": "LM10008"
            }
        ],
        ]
    }
}

这里的 "ones","twos" .. 是动态键，将来还会添加新数组。 所以我们可以在 Flutter 中动态管理它？

Answer 1

我的第一个建议是使用数字而不是字符串来转换这些数据，因此无论将来标签（ones、twos 等）是否发生变化，都不会影响您的代码：

void main() {
  final data = { 
    "success": true, 
    "data": { 
      "ones": [
        { "id": "2", "username": "LM10002" }, 
        { "id": "6", "username": "LM10006" } 
      ], 
      "twos": [
        { "id": "3", "username": "LM10003" }, 
        { "id": "8", "username": "LM10008" } 
      ] 
    }
  };

  print(ApiResponse.fromJson(data));
}

class ApiResponse {
  final bool success;
  final List<List<Data>> data; 
  
  ApiResponse({this.success, this.data});
  
  factory ApiResponse.fromJson(Map<String, dynamic> json) {
    
    List<List<Data>> dataList = [];
    final jsonData = (json["data"] as Map<String, dynamic>);
    jsonData.values.forEach((value) {
      List<Data> parsedList = List<Data>.from(List.from(value).map((element) => Data.fromJson(element)));
      dataList.add(parsedList);
    });
      
    return ApiResponse(
      success: json["success"],
      data: dataList,
    );
  }
  

  @override
  String toString() {
     return "Success: $success, Data: ${data.map((element) => element.map((d) => "${d.id} ${d.username}")).toList().join("; ")}";
  }
}

class Data {
  final String id;
  final String username;
  
  Data({this.id, this.username});
  
  factory Data.fromJson(Map<String, dynamic> json) {
    return Data(
      id: json["id"],
      username: json["username"],
    );
  }
}

这应该返回一个数据列表，其中每个index+1 将代表顺序（一个或两个等...），附加的值将直接是数据列表。

主函数中的打印语句将打印这个：Success: true, Data: (2 LM10002, 6 LM10006); (3 LM10003, 8 LM10008)。

第二种方法是为您的每个数据添加一个group 属性，以便您知道每个数据属于哪个组（ones、twos 等...）

这是完整的实现：

void main() {
  final data = { 
    "success": true, 
    "data": { 
      "ones": [
        { "id": "2", "username": "LM10002" }, 
        { "id": "6", "username": "LM10006" } 
      ], 
      "twos": [
        { "id": "3", "username": "LM10003" }, 
        { "id": "8", "username": "LM10008" } 
      ] 
    }
  };

  print(ApiResponse.fromJson(data));
}

class ApiResponse {
  final bool success;
  final List<Data> data; 
  
  ApiResponse({this.success, this.data});
  
  factory ApiResponse.fromJson(Map<String, dynamic> json) {
    
    List<Data> dataList = [];
    final jsonData = (json["data"] as Map<String, dynamic>);
    jsonData.forEach((key, value) {
      List<Data> parsedList = List<Data>.from(List.from(value).map((element) { 
        element["group"] = key;
        return Data.fromJson(element);
      }));
      dataList.addAll(parsedList);
    });
      
    return ApiResponse(
      success: json["success"],
      data: dataList,
    );
  }

  @override
  String toString() {
     return "Success: $success, Data: ${data.map((element) => "${element.id}, ${element.username}, ${element.group}")}";
  }
}

class Data {
  final String id;
  final String username;
  final String group;
  
  Data({this.id, this.username, this.group});
  
  factory Data.fromJson(Map<String, dynamic> json) {
    return Data(
      id: json["id"],
      username: json["username"],
      group: json["group"]
    );
  }
}

main 方法中的 print 语句应该打印：Success: true, Data: (2, LM10002, ones, 6, LM10006, ones, 3, LM10003, twos, 8, LM10008, twos)

总结一下：

Solution	Description	Advantage	Disadvantage
#1	Proposes to represent the data in the form of a matrix with in ordinates the indexes-1 of the "groups" of data, and in abcissa, the data themselves	We keep the order of the data and it is easy to sort them	Not great for readability, and the group labels returned by the API are
#2	Allows to keep the labels as strings and add them to the data (to later filter the data automatically according to the groups)	We keep the initial data group labels	It will be more difficult to perform sorting

Answer 2

您可以创建一个类来处理这种响应。您将能够访问分配给数据键的映射的不同键并对其执行任何您想要的操作。

class CustomResponse {
  bool success;
  Map<String, dynamic> data;
  
  CustomResponse(this.success, this.data);
  
  factory CustomResponse.fromJson(Map json) {
    return CustomResponse(
      json['success'] as bool,
      json['data'] as Map<String, dynamic>,
    );    
  }
}

void main()  {
  final data = { 
    "success": true, 
    "data": { 
      "ones": [
        { "id": "2", "username": "LM10002" }, 
        { "id": "6", "username": "LM10006" } 
      ], 
      "twos": [
        { "id": "3", "username": "LM10003" }, 
        { "id": "8", "username": "LM10008" } 
      ] 
    }
  };
  
  final customResponse =  CustomResponse.fromJson(data);
  print(customResponse.data.keys);  
}