获取类型“List”不是类型“List”的子类型JSON中的错误

Question

我正在解码一个响应体，我得到了错误：

'List<dynamic>' is not a subtype of type 'List<Example>' 

我正在解析JSON对象的JSON数组，其中一个字段也是对象列表，我怀疑我的问题源于此。我也在使用json_serializable库。下面是我的代码，我省略了一些字段，更改了一些变量名称，但它代表相同的代码：

import 'package:json_annotation/json_annotation.dart';

part 'example_model.g.dart';

@JsonSerializable()
class Example {

  (some fields here)
  final List<Random> some_urls;
  final List<String> file_urls;


  const Example({
    (some fields here)
    this.some_urls,
    this.file_urls,

  });

  factory  Example.fromJson(Map<String, dynamic> json) =>
      _$ ExampleFromJson(json);
}

@JsonSerializable()
class Random {
  final String field_1;
  final int field_2;
  final int field_3;
  final int field_4;
  final bool field_5;

  constRandom(
      {this.field_1, this.field_2, this.field_3, this.field_4, this.field_5});

  factory Random.fromJson(Map<String, dynamic> json) => _$RandomFromJson(json);
}

从。json_serializable生成的g dart文件（将编码部分复制）：

Example _$ExampleFromJson(Map<String, dynamic> json) {
  return Example(

      some_urls: (json['some_urls'] as List)
          ?.map((e) =>
      e == null ? null : Random.fromJson(e as Map<String, dynamic>))
          ?.toList(),
      file_urls: (json['file_urls'] as List)?.map((e) => e as String)?.toList(),

}

Random _$RandomFromJson(Map<String, dynamic> json) {
  return Random(
      field_1: json['field_1'] as String,
      field_2: json['field_2'] as int,
      field_3: json['field_3'] as int,
      field_4: json['field_4'] as int,
      field_5: json['field_5'] as bool);
}

这是我未来的职能：

  Future<List<Example>> getData(int ID, String session) {
    String userID = ID.toString();
    var url = BASE_URL + ":8080/example?userid=${userID}";
    return http.get(url, headers: {
      "Cookie": "characters=${session}"
    }).then((http.Response response) {
      if (response.statusCode == 200) {
        var parsed = json.decode(response.body);
        List<Example> list = parsed.map((i) => Example.fromJson(i)).toList();
        return list;
      }
    }).catchError((e)=>print(e));
  }

Answer 1

我收到了的MappedListIterable

 var parsed = json.decode(response.body);
 var list = parsed.map((i) => Example.fromJson(i)).toList();

将解析的数据转换为List

 var parsed = json.decode(response.body) as List<dynamic>;
 var list = parsed.map((i) => Example.fromJson(i)).toList();

Answer 2

错误原因：

如果源列表的类型为动态或对象（比方说），并且直接将其分配给特定类型而不强制转换，则会出现此错误。

List<dynamic> source = [1]; 
List<int> ints = source; // error

解决方案：

您需要转换列表

1. List<int> ints = List<int>.from(source);
   

   List<int> ints = List.castFrom<dynamic, int>(source);
   

   List<int> ints = source.cast<int>();
   

   List<int> ints = source.map((e) => e as int).toList();
   

Answer 3

此代码创建一个List

parsed.map((i) => Example.fromJson(i)).toList();

必须显式强制转换列表

List<Example> list = List<Example>.from(parsed.map((i) => Example.fromJson(i)));

或者只是

var /* or final */ list = List<Example>.from(parsed.map((i) => Example.fromJson(i)));

另见

• 在Dart中，List.from和. of，Map.from和. of有什么区别？
• https://api.dartlang.org/stable/2.0.0/dart-core/List/List.from.html
• https://api.dartlang.org/stable/2.0.0/dart-core/List/List.of.html
• 飞镖2.X List.cast（）不组成