Ios 自定义嵌套对象的快速编码技术？

我需要获取嵌套对象的编码

class person {
var name: String?
var phone: String?
var address: Address?
}

class Address {
var area: String?
var city: String?
}

我试过了

let data = try? JSONEncoder().encode(person)
let json = try? JSONSerialization.jsonObject(with: data, options: .allowFragments) as! [String: Any]

但地址-键-值对无法获取

struct Person: Encodable {
    var name, phone: String
    var address: Address
}

struct Address: Encodable {
    var area, city: String
}

使您需要编码的模型符合可编码协议 类型应以大写字母开头 如果你确信属性不会为零，不要让它们成为可选的 您可以使模型成为结构而不是类 然后对你的对象进行编码

let data = try? JSONEncoder().encode(person)

当您需要打印编码数据时，需要将其转换为字符串

如果您需要以关键人物和人物对象作为值对对象进行编码，请对字典进行编码

let data = try! JSONEncoder().encode(["person": person])
let json = String(data: data, encoding: .utf8) ?? ""

只需实现可编码

---

您想通过包含缺少的与Person和Address类相关的代码来更新您的问题吗？要使用JSONeCoder对类进行编码，您必须采用Codable。谢谢Robert，但是我需要json{Person:{name:name，phone:phone number，Address:{area:area，city:city name}}}}Person也应该是可编码的。@Satishedy，然后执行encoder.encode[person:person]，其中person是person类型的实例，encoder是JSONEncoder@RobertDresler，因为OP要求在上面的注释中对整个person进行编码。带有所有字段。@user28434 yes，在注释yes中。但问题是，这并不清楚

let data = try! JSONEncoder().encode(["person": person])
let json = String(data: data, encoding: .utf8) ?? ""

class Person: Encodable {
  var name: String?
  var phone: String?
  var address: Address?
}

class Address: Encodable {
  var area: String?
  var city: String?
}
let address = Address()
address.area = "Area"
address.city = "City"
let person = Person()
person.name = "name"
person.address = address

let encoded = try JSONEncoder().encode(person)