如何将 json 字符串传递给对象？答案

【问题标题】：how can I pass a json string to object?如何将 json 字符串传递给对象？
【发布时间】：2019-11-16 11:18:02
【问题描述】：

我的问题是下一个，我有一个这个json

{
  "nombre" : "userProfile.user.name!",
  "apaterno" : 20,
  "amaterno" : true,
  "email" : 100,
  "start_screen" : {
    "info" : true,
    "title" : false,
    "image" : 20,
    "success_btn" : "hola",
    "close_btn" : true
  }
}

我想将此 json 传递给我的结构，我的结构是：

struct person: Decodable
{
    var email : Int
    var nombre : String
    var apaterno : Int
    var amaterno: Bool

    struct start_screen {
        var title: Bool
        var info: Bool
        var image: Int
        var success_btn: String
        var close_btn: Bool
    }
}

我实现的下一行将 json 放入我的结构中，但 start_screen 结构无法获取数据。

let jsonData = json.data(using: .utf8)!
let decoder = JSONDecoder()
let myStruct = try! decoder.decode(person.self, from: jsonData)

当我访问 myStruct.email 时，我得到 100，没问题，但我无法加载 start_screen 数据，我该怎么办？

【问题讨论】：

标签： json swift dictionary object struct

【解决方案1】：

这应该是你的 Person 结构：

struct Person: Decodable {
var email : Int?
var nombre : String?
var apaterno : Int?
var amaterno: Bool
var start_screen: Start_screen?

enum CodingKeys: String, CodingKey {

    case email = "email"
    case nombre = "nombre"
    case apaterno = "apaterno"
    case amaterno = "amaterno"
    case start_screen = "amaterno"
}

init(from decoder: Decoder) throws {
    let values = try decoder.container(keyedBy: CodingKeys.self)
    email = try values.decodeIfPresent(Int.self, forKey: .email)
    nombre = try values.decodeIfPresent(String.self, forKey: .nombre)
    apaterno = try values.decodeIfPresent(Int.self, forKey: .apaterno)
    amaterno = try values.decodeIfPresent(Bool.self, forKey: .apaterno) ?? false
    start_screen = try values.decodeIfPresent(Start_screen.self, forKey: .start_screen)
}

}

这应该是你的 Start_screen 结构：

struct Start_screen: Decodable {
var title: Bool
var info: Bool
var image: Int?
var success_btn: String?
var close_btn: Bool

enum CodingKeys: String, CodingKey {

    case title = "title"
    case info = "info"
    case image = "image"
    case success_btn = "success_btn"
    case close_btn = "close_btn"
}

init(from decoder: Decoder) throws {
    let values = try decoder.container(keyedBy: CodingKeys.self)
    title = try values.decodeIfPresent(Bool.self, forKey: .title) ?? false
    info = try values.decodeIfPresent(Bool.self, forKey: .info) ?? false
    image = try values.decodeIfPresent(Int.self, forKey: .image)
    success_btn = try values.decodeIfPresent(String.self, forKey: .success_btn)
    close_btn = try values.decodeIfPresent(Bool.self, forKey: .close_btn) ?? false
}

}

从Person访问start_screen：

if let jsonData = json.data(using: .utf8) {
   let user = try! JSONDecoder().decode(Person.self, from: jsonData)
   if let title = user.start_screen.title {
      print(title)
   }
}

【讨论】：

【解决方案2】：

首先，您需要为start_screen 添加一个变量到person。

var start_screen: start_screen

那么你需要start_screenDecodable

struct start_screen: Decodable

这应该是让它工作的最小更改量。

此外，您可能希望将您的类型大写。 start_screen: start_screen 看起来真的很混乱。您还可以将变量和类型名称设为camelCase，并让JSONDecoder 转换为/从snake_case 为您转换。这也是swift中的命名约定。它看起来像这样

struct Person: Decodable {
    var email: Int
    var nombre: String
    var apaterno: Int
    var amaterno: Bool
    var startScreen: StartScreen

    struct StartScreen: Decodable {
        var title: Bool
        var info: Bool
        var image: Int
        var successBtn: String
        var closeBtn: Bool
    }
}

let jsonData = json.data(using: .utf8)!
let decoder = JSONDecoder()
decoder.keyDecodingStrategy = .convertFromSnakeCase
let person = try! decoder.decode(Person.self, from: jsonData)
print(person)

【讨论】：