firebase - 如何在 flutter 应用程序的 Cloud firestore 中记录带有嵌套数组的 json

Question

我在 flutter 应用程序中使用了一个 json 文件，但在 firestore 中记录这个 json 失败了。

我的 json 在 jsonCloudContent 中。以下不起作用:

Firestore.instance.collection('toto').document().setData(jsonCloudContent);

当我尝试使用 json.encode(jsonCloudContent) 中的字符串时，它工作正常:

Firestore.instance.collection('userStuffLists').document().setData({"apiVersion":2,"userId":"unknown","googleId":"unknown","list":[{"userI...

这是一个 json 文件的例子:https://jsoneditoronline.org/?id=3ac25ef1743047e08467cccbad031d5e

上传数据的函数:

Future<bool> uploadUserStuffList() async {
  Map<String, dynamic> jsonCloudContent = StuffList(
    list: list,
    apiVersion: apiVersion,
    userId: userId,
    googleId: googleId,
    listVersion: listVersion)
    .toJson();
  try {
    Firestore.instance.collection('toto').document().setData(jsonCloudContent);
  } catch (e) {
    debugPrint(e);
  }
    return true;
  }

在 Intellij inspector 中，格式看起来不错，但为了以防万一，这里是对象定义:

class Stuff {
  final String description;
  final List<String> imagePath;
  final String userId;
// TODO : put unknown in the key_strings ?
  Stuff({this.userId = 'unknown', this.description, this.imagePath});

  factory Stuff.fromJson(Map<String, dynamic> json) {
    return new Stuff(
      userId: json['userId'] != null ? json['userId'] : 'unknown',
//      TODO : throw error if description is null ? not certain
      description: json['description'],
      imagePath: json['path'] != null ? List<String>.from(json['path']) : null,
    );
  }

  Map<String, dynamic> toJson() => {
        'userId': userId != null ? userId : 'unknown',
        'description': description,
        'path': imagePath,
      };
}
class StuffList {
  final List<Stuff> list;
  final int apiVersion;
//  TODO need appropriate format when integrate firebase and google authentication
  final String userId;
  final String googleId;
  final int listVersion;
  StuffList({
    this.apiVersion = 2,
    this.userId = 'unknown',
    this.googleId,
    this.listVersion = 1,
    this.list,
  });

  factory StuffList.fromJson(Map<String, dynamic> json) {
    List<Stuff> listOfStuff = [];
    if (json['list'] != null) {
      json['list'].forEach((content) {
        Stuff stuff = Stuff.fromJson(content);
        listOfStuff.add(stuff);
      });
    } else {
      listOfStuff = null;
    }

    return new StuffList(
      apiVersion: json['apiVersion'],
      userId: json['userId'] != null ? json['userId'] : 'unknown',
      googleId: json['googleId'] != null ? json['googleId'] : 'unknown',
      listVersion: json['listVersion'],
//      list: json['list'] != null ? (json['list'] as List).map((i) => new Stuff.fromJson(i)) : null,
      list: listOfStuff,
    );
  }

  Map<String, dynamic> toJson() => {
        'apiVersion': apiVersion != null ? apiVersion : 1,
        'userId': userId != null ? userId : 'unknown',
        'googleId': googleId != null ? googleId : 'unknown',
        'listVersion': listVersion != null ? listVersion : 1,
        'list': list,
      };
}

Answer 1

发现我的错误。对象 StuffList 包含 Stuff 对象的列表。toJson 方法没有将 Stuff 列表转换为 json。这是新方法:

  Map<String, dynamic> toJson() => {
        'apiVersion': apiVersion != null ? apiVersion : 1,
        'userId': userId != null ? userId : 'unknown',
        'googleId': googleId != null ? googleId : 'unknown',
        'listVersion': listVersion != null ? listVersion : 1,
        'list': list.map((i) => i.toJson()).toList(),
      };