從API獲取資料時出現錯誤，如果出現抖動，我應該如何獲取資料？

我正在API使用模型獲取資料。但是我遇到了一個問題，就是在獲取資料的時候'gallery'報錯，就是我獲取的資料不正確。我需要獲取該'gallery'欄位并在其中獲取該'url'欄位-照片的鏈接，以便將來使用它。您能告訴我如何'url'正確獲取該欄位嗎？

{
    "data": {
        "id": 35,
        "picture_url": null,
        "email_confirmed": false,
        "gallery": [
            {
                "url": "https://picture-staging.s3.eu-central.jpeg",
                "mime_type": "image/jpeg",
                "type": "gallery",
                "updated_at": "2022",
                "created_at": "2022"
            }
        ],
        "updated_at": "2022",
        "created_at": "2022"
    }
}

模型

class User {
  final int id;
  List? gallery;

  User({
    required this.id,
    this.gallery,
  });

  User.fromJson(Map<String, dynamic> json)
      : this(
          id: json['id'] as int,
          gallery: json['gallery']['url'],
        );

uj5u.com熱心網友回復：

在您的 API 回應中，有一個畫廊物件串列，因此您必須遍歷所有這些物件。

User.fromJson(Map<String, dynamic> json) {
    json = json['data'];
    id = json['id'];
    pictureUrl = json['picture_url'];
    emailConfirmed = json['email_confirmed'];
    if (json['gallery'] != null) {
      gallery = <Gallery>[];
      json['gallery'].forEach((v) {
        gallery!.add(new Gallery.fromJson(v));
      });
    }
    updatedAt = json['updated_at'];
    createdAt = json['created_at'];
  }

有多種工具可以幫助您創建 .fromJson 方法，例如. 將你的 json 粘貼到那里，它會為你生成 dart 代碼，這對我很有幫助。

用法應該是這樣的：

User user = User.fromJson(yourApiResponseJson);
print(user.id);
print(user.gallery); //prints entire list of gallery
print(user.gallery.first.url); //prints only first object url

uj5u.com熱心網友回復：

我希望這不是您的整個模型，因為該模型沒有訪問"data"json 回應上的密鑰，您的模型應該開始獲取密鑰data然后將其傳遞給另一個類，在這種情況下應該命名User

這是一個簡單的例子

class User {
    User({
        required this.data,
    });

    final Data data;

    factory User.fromJson(Map<String, dynamic> json) => User(
        data: Data.fromJson(json["data"]),
    );
}

Data類可能是這樣的：

class Data {
    Data({
        required this.id,
        required this.pictureUrl,
        required this.emailConfirmed,
        required this.gallery,
        required this.updatedAt,
        required this.createdAt,
    });

    final int id;
    final dynamic pictureUrl;
    final bool emailConfirmed;
    final List<Gallery> gallery;
    final String updatedAt;
    final String createdAt;

    factory Data.fromJson(Map<String, dynamic> json) => Data(
        id: json["id"],
        pictureUrl: json["picture_url"],
        emailConfirmed: json["email_confirmed"],
        gallery: List<Gallery>.from(json["gallery"].map((x) => Gallery.fromJson(x))),
        updatedAt: json["updated_at"],
        createdAt: json["created_at"],
    );
    
}

我推薦你使用Quicktype

uj5u.com熱心網友回復：

嘿，您可以使用此工具從 json 生成您的飛鏢模型。

下面是從上面的工具生成的代碼

// final user = userFromJson(jsonString);
import 'dart:convert';

User userFromJson(String str) => User.fromJson(json.decode(str));

String userToJson(User data) => json.encode(data.toJson());

class User {
  User({
    required this.data,
  });

  Data data;

  factory User.fromJson(Map<String, dynamic> json) => User(
    data: Data.fromJson(json["data"]),
  );

  Map<String, dynamic> toJson() => {
    "data": data.toJson(),
  };
}

class Data {
  Data({
    this.id,
    this.pictureUrl,
    this.emailConfirmed,
    this.gallery,
    this.updatedAt,
    this.createdAt,
  });

  int? id;
  String? pictureUrl;
  bool? emailConfirmed;
  List<Gallery>? gallery;
  String? updatedAt;
  String? createdAt;

  factory Data.fromJson(Map<String, dynamic> json) => Data(
    id: json["id"],
    pictureUrl: json["picture_url"],
    emailConfirmed: json["email_confirmed"],
    gallery: List<Gallery>.from(json["gallery"].map((x) => Gallery.fromJson(x))),
    updatedAt: json["updated_at"],
    createdAt: json["created_at"],
  );

  Map<String, dynamic> toJson() => {
    "id": id,
    "picture_url": pictureUrl,
    "email_confirmed": emailConfirmed,
    "gallery": List<dynamic>.from(gallery.map((x) => x.toJson())),
    "updated_at": updatedAt,
    "created_at": createdAt,
  };
}

class Gallery {
  Gallery({
    this.url,
    this.mimeType,
    this.type,
    this.updatedAt,
    this.createdAt,
  });

  String? url;
  String? mimeType;
  String? type;
  String? updatedAt;
  String? createdAt;

  factory Gallery.fromJson(Map<String, dynamic> json) => Gallery(
    url: json["url"],
    mimeType: json["mime_type"],
    type: json["type"],
    updatedAt: json["updated_at"],
    createdAt: json["created_at"],
  );

  Map<String, dynamic> toJson() => {
    "url": url,
    "mime_type": mimeType,
    "type": type,
    "updated_at": updatedAt,
    "created_at": createdAt,
  };
}

// 你可以這樣使用

 final user = userFromJson(jsonString);
 String? url =  user.data?.gallery?.url;

標籤：扑 api 镖

上一篇：如何檢查Python的APIGET請求中是否存在特定的JSON值？

下一篇：如何使用GitRESTAPI從頁面中提取所有repo