获取类型 'List<dynamic>'不是“List<...>"类型的子类型JSON 中的错误 [英] Getting type &#39;List&lt;dynamic&gt;&#39; is not a subtype of type &#39;List&lt;...&gt;&#39; error in JSON

问题描述

我正在解码响应正文，但出现错误:

I'm decoding a response body and I'm getting the error:

'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，我省略了一些字段并更改了一些变量名称，但它表示相同的代码:

I'm parsing a JSON array of JSON objects, one of the fields is a list of objects as well and I suspect my issue stems from that. I am also using the json_serializable library. Below is my code, I omitted some fields out and changed some variable names but it represents the same code:

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);
}

来自 json_serializable 生成的 .g dart 文件(省略了编码部分):

from the .g dart file that json_serializable made (ommited the encoding part):

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);
}

这是我未来的功能:

  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));
  }

推荐答案

这段代码创建了一个List

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

您必须像这样将 List 显式转换为 List，

You must explicitly cast List<dynamic> to List<Example> like so,

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 以及 Map.from 和 .of 之间有什么区别?
• 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
• Dart 2.X List.cast() 不组合

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