iOS：如何对具有协议类型的变量的模型类使用可解码的代码

Question

问题：有一个符合Decodable的模型类，该模型的变量类型为someProtocol。但是编译器给出一个错误，

编译器错误

Type 'MyModel' does not conform to protocol 'Decodable'
Protocol requires initializer 'init(from:)' with type 'Decodable'
Cannot automatically synthesize 'Decodable' because 'SomeProtocol' does not conform to 
'Decodable'

代码段

class MyModel: Decodable {
var name: String?
var employee: SomeProtocol?

enum CodingKeys: String, CodingKey {
    case name
    case employee
}

} // Enf of class MyModel

protocol SomeProtocol {
 var employeeName: String ? { get }
}

Answer 1

[已解决]嗨，所以我想这个问题是有效的，经过一番搜索，我能够解决。MyModel类不能具有协议类型，应该有一个继承协议的具体类型类。现在，对象也与嵌套的其他模型对象一起被解码。

更正上面的代码

class MyModel: Decodable {
var name: String?
var employee: OtherModel?

enum CodingKeys: String, CodingKey {
case name = "DepartmentName"
case employee
}

} // End of class MyModel

class OtherModel: SomeProtocol {
var employeeName: String?
}

protocol SomeProtocol: Decodable {
 var employeeName: String? { get }
}

注：不需要init（来自解码器：Decoder）在这种情况下。