一尘不染

获取类型“列表'不是JSON中错误'List <...>'类型的子类型

flutter

我正在解码响应正文,但出现错误:

'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,并且我怀疑我的问题是由此引起的。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,但是它表示相同的代码:

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);
}

来自json_serializable制作的.g dart文件(省略了编码部分):

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);
}

这是我将来的功能:

  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));
  }

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2020-08-13

共1个答案

一尘不染

此代码创建了一个 List<dynamic>

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

用代替

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

要不就

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

也可以看看

2020-08-13