列表中的 Dart 对象答案

【问题标题】：Dart objects in Lists列表中的 Dart 对象
【发布时间】：2021-03-30 04:54:15
【问题描述】：

我是 dart/flutter 的新手。

如何使用列表中的对象？

我有这样的对象：

{
    "channelName": "sydneyfunnelaio",
    "type": "",
    "ChannelPic": "https://static-cdn.jtvnw.net/jtv_user_pictures/8ead1810-f82a-4dc0-a3a6-583171baff60-profile_image-300x300.jpeg",
    "success": true
}

我怎样才能用它创建列表/数组；

我想喜欢：

[{
    "channelName": "sydneyfunnelaio",
    "type": "",
    "ChannelPic": "https://static-cdn.jtvnw.net/jtv_user_pictures/8ead1810-f82a-4dc0-a3a6-583171baff60-profile_image-300x300.jpeg",
    "success": true
},{
    "channelName": "qweqdqaw",
    "type": "",
    "ChannelPic": "https://static-cdn.jtvnw.net/jtv_user_pictures/8ead1810-f82a-4dc0-a3a6-583171baff60-profile_image-300x300.jpeg",
    "success": true
}]

【问题讨论】：

标签： arrays list flutter dart

【解决方案1】：

你可以试试这样的：

void main() {
  List<MyObject> myObjects = [];

  myObjects.add(MyObject.fromJson({
    "channelName": "sydneyfunnelaio",
    "type": "",
    "ChannelPic":
        "https://static-cdn.jtvnw.net/jtv_user_pictures/8ead1810-f82a-4dc0-a3a6-583171baff60-profile_image-300x300.jpeg",
    "success": true
  }));

  myObjects.add(MyObject.fromJson({
    "channelName": "qweqdqaw",
    "type": "",
    "ChannelPic":
        "https://static-cdn.jtvnw.net/jtv_user_pictures/8ead1810-f82a-4dc0-a3a6-583171baff60-profile_image-300x300.jpeg",
    "success": true
  }));

  print(myObjects);
  print(myObjects[0].channelName);
  print(myObjects[1].channelName);

  myObjects.forEach((obj)=>print(obj.toJson()));

}

class MyObject {
  String channelName;
  String type;
  String channelPic;
  bool success;
  MyObject({this.channelName, this.type, this.channelPic, this.success});
  MyObject.fromJson(Map<String, dynamic> json) {
    channelName = json['channelName'];
    type = json['type'];
    channelPic = json['ChannelPic'];
    success = json['success'];
  }
  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = new Map<String, dynamic>();
    data['channelName'] = this.channelName;
    data['type'] = this.type;
    data['ChannelPic'] = this.channelPic;
    data['success'] = this.success;
    return data;
  }
}

【讨论】：