Json可序列化为 Json 抛出 return 错误

Question

我正在尝试使用这个包 - https://pub.dev/packages/json_serializable 来帮助我解析来自 api 的数据。但是我遇到了下面的错误，我不知道如何解决它。

我收到这个错误 A value of type 'Location' can't be returned from the method 'toJson' because it has a return type of 'Map<String, dynamic>

@JsonSerializable()
class Location {
  const Location({
    required this.title,
    required this.locationType,
    required this.latLng,
    required this.woeid,
  });

  final String title;
  final LocationType locationType;
  final LatLng latLng;
  final int woeid;

  factory Location.fromJson(Map<String,dynamic> json) => _$LocationFromJson(json);
  Map<String, dynamic> toJson() => _$LocationFromJson(this);
}

每当我将最后一部分更改为 Location toJson() => _$LocationFromJson(this); 时，我都会收到错误 The argument type 'Location' can't be assigned to the parameter type 'Map<String, dynamic>

Answer 1

您需要将复杂的数据类型转换为简单的数据类型，如String、int、double等，以便在json中进行解析。下面是一个关于如何实现纬度和经度的示例。

假设我有一个学校数据，我想为其存储名称和位置。我会为名称创建一个简单的字符串，但为位置创建一个单独的 class。以下是它的运行方式：

class SchoolDataModel {
  final String name;
  final SchoolLocation schoolLocation;

  SchoolDataModel({
    required this.name,
    required this.schoolLocation,
  });

  factory SchoolDataModel.fromJson(Map<String, dynamic> parsedJson) =>
      SchoolDataModel(
        name: parsedJson['name'] as String,
        schoolLocation: SchoolLocation.fromJson(
            parsedJson['location'] as Map<String, dynamic>),
      );
}

如您所见，SchoolLocation.fromJson用于使用自己的构造函数解析位置数据。如下所示：

class SchoolLocation {
  final String latitude;
  final String longitude;

  SchoolLocation({
    required this.latitude,
    required this.longitude,
  });

  factory SchoolLocation.fromJson(Map<String, dynamic> parsedJson) =>
      SchoolLocation(
        latitude: parsedJson['latitude'] as String,
        longitude: parsedJson['longitude'] as String,
      );
}

Json可序列化为 Json 抛出 return 错误

JsonSerializable to Json throws return error

dart

flutter