铸造模型不适用于 getX 颤振

Posted 2023-03-07

【中文标题】铸造模型不适用于 getX 颤振

【英文标题】：cast in model not working wit getX flutter

【发布时间】：2021-09-28 17:11:39

【问题描述】：

我有一个来自 https://newsapi.org/ 的带有新闻 api 的应用程序

我的 quicktype 模型：

// To parse this JSON data, do
//
//     final news = newsFromJson(jsonString);

import 'package:meta/meta.dart';
import 'dart:convert';

List<News> newsFromJson(String str) =>
    List<News>.from(json.decode(str).map((x) => News.fromJson(x)));

String newsToJson(List<News> data) =>
    json.encode(List<dynamic>.from(data.map((x) => x.toJson())));

class News 
  News(
    required this.status,
    required this.totalResults,
    required this.articles,
  );

  final String status;
  final int totalResults;
  final List<Article> articles;

  factory News.fromJson(Map<String, dynamic> json) => News(
        status: json["status"],
        totalResults: json["totalResults"],
        articles: List<Article>.from(
            json["articles"].map((x) => Article.fromJson(x))),
      );

  Map<String, dynamic> toJson() => 
        "status": status,
        "totalResults": totalResults,
        "articles": List<dynamic>.from(articles.map((x) => x.toJson())),
      ;


class Article 
  Article(
    required this.source,
    required this.author,
    required this.title,
    required this.description,
    required this.url,
    required this.urlToImage,
    required this.publishedAt,
    required this.content,
  );

  final Source source;
  final String author;
  final String title;
  final String description;
  final String url;
  final String urlToImage;
  final DateTime publishedAt;
  final String content;

  factory Article.fromJson(Map<String, dynamic> json) => Article(
        source: Source.fromJson(json["source"]),
        author: json["author"] == null ? null : json["author"],
        title: json["title"],
        description: json["description"],
        url: json["url"],
        urlToImage: json["urlToImage"],
        publishedAt: DateTime.parse(json["publishedAt"]),
        content: json["content"],
      );

  Map<String, dynamic> toJson() => 
        "source": source.toJson(),
        "author": author == null ? null : author,
        "title": title,
        "description": description,
        "url": url,
        "urlToImage": urlToImage,
        "publishedAt": publishedAt.toIso8601String(),
        "content": content,
      ;


class Source 
  Source(
    required this.id,
    required this.name,
  );

  final String id;
  final String name;

  factory Source.fromJson(Map<String, dynamic> json) => Source(
        id: json["id"] == null ? null : json["id"],
        name: json["name"],
      );

  Map<String, dynamic> toJson() => 
        "id": id == null ? null : id,
        "name": name,
      ;

在我的 remoteservice.dart 中：

import 'package:http/http.dart' as http;
import 'package:nocovid/models/news.dart';
import 'package:nocovid/utils/constant.dart';

class RemoteServices 
  static var client = http.Client();

  static Future<List<News>?> fetchNews() async 
    final String endpoint =
        'https://newsapi.org/v2/everything?q=covid19&apiKey=' + kAPIKey;
    final Uri url = Uri.parse(endpoint);

    final response = await client.get(url);
    if (response.statusCode == 200) 
      var jsonString = response.body;
      return newsFromJson(jsonString);
     else 
      return null;
    
  

newscontroller.dart

import 'package:get/state_manager.dart';
import 'package:nocovid/models/news.dart';
import 'package:nocovid/services/remote_services.dart';

class NewsController extends GetxController 
  var newsList = <News>[].obs;

  @override
  void onInit() 
    fetchNews();
    super.onInit();
  

  void fetchNews() async 
    var news = await RemoteServices.fetchNews();
    if (news != null) 
      newsList.value = news;
    
  

并得到这个错误： 和

调用会定期执行，但在显示数据时会生成这些错误。 我检查了 github 上的一些代码，一切似乎都正常，但我无法继续

【参考方案1】：

改变

List<News> newsFromJson(String str) =>
    List<News>.from(json.decode(str).map((x) => News.fromJson(x)));

到

News newsFromJson(String str) => News.fromJson(json.decode(str));

原因是 News 对象不是一个列表，它是一个复杂的 JSON，其中包含一个由文章列表组成的映射。您需要正确通过API。

如果你想要一个模型，你可以使用quicktype。只需粘贴 URL 响应即可。

也改变

 static Future<List<News>?> fetchNews() 

到

 static Future<News> fetchNews() 

【讨论】：

无法从方法“fetchNews”返回类型为“News”的值，因为它的返回类型为“Future?>”

我已经编辑了我的答案。