从API获取数据时，我遇到了一个错误，如果颤抖，我应该如何获取数据？

Question

我使用模型从api获取数据。但是，我遇到了一个问题，即当我获得'Gallery'数据时，我会遇到一个错误，也就是说，我得到了错误的数据。我需要获取'gallery'字段，并在其内部使用'url'字段 - 照片的链接，以便将来使用它。您能告诉我如何正确获取'url'字段？

{
    "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'],
        );

Answer 1

在您的API响应中，有一个画廊对象列表，因此您必须穿越所有内容。

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'];
  }

有多种工具可以帮助您创建该.fromjson方法，例如 this 。在那里粘贴您的JSON，它将为您生成Dart代码，真的对我有帮助。

用法应该这样：

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

Answer 2

我希望这不是您的整个模型，因为该模型无法访问“数据” json响应中的密钥，您的模型应开始获取键data然后传递对于在这种情况下应命名用户的另一个类

，这是一个简短的示例

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

    final Data data;

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

data类可能是这样的：

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"],
    );
    
}

我建议您使用 quickType

Answer 3

嘿，您可以使用此工具从JSON生成DART模型。

以下是从上方生成的代码

// 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,
  };
}

//您可以这样使用

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