解析失败: JSONDecoder Swift

Question

我正在尝试用AlamoFire5.2解码一个json请求，问题是我使用了JSONDecoder，并且我对转换有一些问题

API是西班牙语的，我的模型是英语的，所以我决定使用键枚举来更改这种值。

但我不知道这是否有效..。这是我的密码：

API响应：(json变量)

    {
  "sistemaOperativoId" : 0,
  "nombreUsuario" : "Coasnf_09",
  "menus" : [

  ],
  "acciones" : [

  ],
  "fechaRegistro" : "2020-04-15T09:46:24.0573154",
  "empresa" : null,
  "version" : null
}

我的模特：

struct UserP: Decodable{
    var username : String
    var company : String

    private enum CodingKeys: String, CodingKey{
        case username = "nombreUsuario"
        case company = "empresa"
    }

    init(from decoder: Decoder) throws{
        let container = try decoder.container(keyedBy: CodingKeys.self)
        username = try container.decode(String.self, forKey: .username) ?? "null"
        company = try container.decode(String.self, forKey: .company)  ?? "null"
    }

    init(username: String, company: String){
        self.username = username
        self.company = company
    }
}

协调：

   func login(user: User) -> UserP? {
        var userData: UserP! = nil
        AF.request(UserRouter.login(user: user)).responseJSON{ response in
            switch response.result {
            case .success(let response):
                print(response)
                let dict = (response as? [String : Any])!
                let json = dict["data"] as! [String: Any]

                if let jsonData = try? JSONSerialization.data(withJSONObject: json , options: .prettyPrinted)
                {
                    do {
                        var jsonString = String(data: jsonData, encoding: String.Encoding.utf8)!
                        print(jsonString)
                        userData = try JSONDecoder().decode(UserP.self, from: Data(jsonString.utf8))
                        print("Object Converted: ", userData.username)
                    } catch {
                        print("Parsing Failed: ", error.localizedDescription)
                    }
                }



                break

            case .failure(let error):
                print(error)
                break
            }
        }

        return userData
    }

Answer 1

与阿拉莫火逻辑有关的变化：

response from Alamofire有一个data属性，可以直接使用：

let json = response.data
do {
    let user = try JSONDecoder().decode(UserP.self, from: json)
    print(user)
}
catch {
    print(error)
}

如果键nombreUsuario和empresa不能保证在json中出现，那么推荐的方法是将这些变量标记为可选的。这样您就不需要定制的解码器逻辑了。

模型更改#1：

struct UserP: Decodable {
    var username: String?
    var company: String?

    private enum CodingKeys: String, CodingKey{
        case username = "nombreUsuario"
        case company = "empresa"
    }
}

模型更改#2：

如果您想给出一些默认值，那么自定义解码器可以帮助：

struct UserP: Decodable {
    var username: String
    var company: String

    private enum CodingKeys: String, CodingKey{
        case username = "nombreUsuario"
        case company = "empresa"
    }

    init(from decoder: Decoder) throws{
        let container = try decoder.container(keyedBy: CodingKeys.self)
        username = try container.decodeIfPresent(String.self, forKey: .username) ?? "Default Name"
        company = try container.decodeIfPresent(String.self, forKey: .company) ?? "Default Company Name"
    }
}