如何使用嵌套的 map 字段为带有 flutter 的云 Firestore 创建用户 model？

Question

我想创建一个类型为 map 的嵌套字段。 'usersName'是类型为 map 的字段，包含'firstName'和'lastName' 。 这是图像： 如果你需要源代码，请告诉我。 谢谢

用户模型.dart

class User {
  String id;
  String name;
  String email;
  String password;
  String phone;

  User({
    required this.id,
    required this.name,
    required this.email,
    required this.password,
    required this.phone,
  });

  //and i dont know how to make it work
}

Answer 1

进行此更改，您将能够根据需要接收它：

class User {
  String id;
  Map<String, String> name;
  String email;
  String password;
  String phone;

  User({
    required this.id,
    required this.name,
    required this.email,
    required this.password,
    required this.phone,
  });

  //and i dont know how to make it work
}

有关 dart 地图的更多信息： https://dart.dev/guides/language/language-tour#maps

Answer 2

您可以使用firstName和lastName变量创建一个UserName class ，然后在您的User class 上将您的name从String更改为UserName 。

class User {
 String id;
 UserName name;
 String email;
 String password;
 String phone;

 User({
   required this.id,
   required this.name,
   required this.email,
   required this.password,
   required this.phone,
 });
}

class UserName {
   String firstName;
   String lastName;
   UserName({
      required this.firstName,
      required this.lastName,
   });
}

Answer 3

我同意将name属性设置为Map是通往 go 的方式。例如，您的用户名如下：

var name = {firstName: "Joe", lastName: "Smith"}

如上所述创建您的User class 然后创建方法来相应地转换您的属性。 在这里，我创建了toMap()以返回用户的 Map object。

class User {
  String id;
  Map<String, String> name;
  String email;
  String password;
  String phone;

  User({
    required this.id,
    required this.name,
    required this.email,
    required this.password,
    required this.phone,
  });

  // if you intend to store this User object into a database/storage
  // that requires Map object
   Map<String, dynamic> toMap() {
    // convert Map to String
    var fullName = name.toString();
  
    return {
      'id': id,
      // String key, String value
      'name': fullName,
      ...
    };
  }

}