Flutter中如何将Firebase检索到的数据正确追加到列表返回？

我正尝试编写一个函数，用于从Firebase Realtime Database获取并返回文章列表，但该功能无法正常运行，我是Flutter开发新手。
我的数据库结构如下所示：

我编写的实现代码如下：

Future<List<ArticleModel>> getArticles() async {
  List<ArticleModel> articles = [];
  uid = FirebaseAuth.instance.currentUser!.uid;

  var ref = FirebaseDatabase.instance.ref().child("saved").child(uid);
  await ref.once().then((DatabaseEvent databaseEvent) {
    var docs = databaseEvent.snapshot.children;
    for (var element in docs) {
      Map<dynamic, dynamic> data = jsonDecode(jsonEncode(element.value));
      var article = ArticleModel.fromJson(Map<dynamic, dynamic>.from(data));
      articles.add(article);
    }
  });
  print(articles);//此处打印结果为[Instance of 'ArticleModel', Instance of 'ArticleModel'.......
  return articles;
}

ArticleModel类的实现代码如下：

class ArticleModel {
  String sourceName;
  String author;
  String title;
  String description;
  String url;
  String urlToImage;
  String publishedAt;

  ArticleModel(
      {required this.sourceName,
      required this.author,
      required this.title,
      required this.description,
      required this.urlToImage,
      required this.url,
      required this.publishedAt});

  factory ArticleModel.fromJson(Map<dynamic, dynamic> element) {
    return ArticleModel(
        author: element['author'],
        description: element['desc'],
        sourceName: element['source'],
        publishedAt: element['time'],
        title: element['title'],
        url: element['url'],
        urlToImage: element['urlImage'],
  }
}

现有代码存在4个核心问题：

• ArticleModel.fromJson工厂方法缺少闭合括号，属于语法错误，会直接导致编译不通过
• 打印结果显示Instance of 'ArticleModel'是Dart默认行为，不是数据获取失败：Dart打印类实例时默认不会输出实例内部属性，只输出类型标识
• 对Firebase返回的快照值做jsonDecode(jsonEncode())二次转换属于冗余操作，Firebase返回的snapshot.value本身就是标准Map结构，多余的编解码反而可能触发类型转换异常
• 没有做空值兜底，当节点下无数据、或者某条文章缺少个别字段时，会直接抛出空指针错误

1. 修正ArticleModel类，补全语法、增加空值兼容、重写toString方法方便调试 #

class ArticleModel {
  String sourceName;
  String author;
  String title;
  String description;
  String url;
  String urlToImage;
  String publishedAt;

  ArticleModel(
      {required this.sourceName,
      required this.author,
      required this.title,
      required this.description,
      required this.urlToImage,
      required this.url,
      required this.publishedAt});

  // 重写toString方法，打印时直接显示属性内容
  @override
  String toString() {
    return 'ArticleModel{sourceName: $sourceName, author: $author, title: $title, description: $description, url: $url, urlToImage: $urlToImage, publishedAt: $publishedAt}';
  }

  factory ArticleModel.fromJson(Map<dynamic, dynamic> element) {
    // 增加空值兜底，避免数据库字段缺失导致崩溃
    return ArticleModel(
        author: element['author'] ?? '',
        description: element['desc'] ?? '',
        sourceName: element['source'] ?? '',
        publishedAt: element['time'] ?? '',
        title: element['title'] ?? '',
        url: element['url'] ?? '',
        urlToImage: element['urlImage'] ?? ''
    ); // 补全之前缺失的右括号
  }
}

2. 优化数据获取函数，去掉冗余转换、增加空判断 #

Future<List<ArticleModel>> getArticles() async {
  List<ArticleModel> articles = [];
  final uid = FirebaseAuth.instance.currentUser!.uid;

  final ref = FirebaseDatabase.instance.ref().child("saved").child(uid);
  final databaseEvent = await ref.once();
  // 节点无数据时直接返回空列表
  if (databaseEvent.snapshot.value != null) {
    final docs = databaseEvent.snapshot.children;
    for (var element in docs) {
      // 直接转Map即可，不需要多余的json编解码
      final data = Map<dynamic, dynamic>.from(element.value as Map);
      articles.add(ArticleModel.fromJson(data));
    }
  }
  print(articles); // 重写toString后这里会打印完整的文章属性
  return articles;
}

3. 调用注意事项 #

getArticles()是异步方法，调用时必须加await，否则拿不到实际返回的列表：

// 正确调用示例
final articleList = await getArticles();

请核对数据库实际字段名和fromJson中读取的key完全匹配，若数据库存储的字段名和代码中写的desc/source/time/urlImage不一致，对应修改即可，否则会读取到空值。

内容的提问来源于stack exchange，提问作者Anas Ansari