JsonSerializabletoJson拋出回傳錯誤-有解無憂

我正在嘗試使用這個包 - https://pub.dev/packages/json_serializable來幫助我決議來自 api 的資料。但是我面臨以下錯誤，我不知道如何解決它。

我收到這個錯誤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);
}

每當我將最后一部分更改為Location toJson() => _$LocationFromJson(this);然后我得到錯誤The argument type 'Location' can't be assigned to the parameter type 'Map<String, dynamic>

uj5u.com熱心網友回復：

您需要將復雜的資料型別轉換為簡單的資料型別，如String、int、double等，以便在json中決議。這是有關如何實作緯度和經度的示例。

假設我有一個學校資料，我想為其存盤名稱和位置。我會為名稱創建一個簡單的字串，但為位置創建一個單獨的類。這是怎么回事：

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>),
      );
}

如您所見，SchoolLocation.fromJson 用于使用自己的建構式決議位置資料。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,
      );
}

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