问题描述
Swift作为Codable
(Decodable
& Encodable
)协议的这种功能非常有用.但是我发现了这样的问题:让我们让类Parent符合Codable
:
Such feature of Swift as Codable
(Decodable
&Encodable
) protocol is very useful.But I found such issue:Let we have class Parent conforming to Codable
:
class Parent: Codable {
var name: String
var email: String?
var password: String?
}
好吧,该类符合从盒子开始"的Codable协议,您无需编写任何初始化程序,就可以像这样从JSON进行初始化:
Ok, that class is conforming to Codable protocol "from the box", you don't need write any initializers, it's ready to be initialized from JSON like that:
{ "name": "John", "email": "[email protected]", "password": <null>}
但是假设我们需要其他类,Child从Parent继承并符合Codable:
But let's say we need other class, Child inherits from Parent and be conforming to Codable:
class Child: Parent {
var token: String
var date: Date?
}
因此,子类别必须通过与父项相符合来使之与Codable相符合,无法从JSON正确初始化Child类的BUT属性.我发现的决定是由我自己编写Child类的所有Codable内容,例如:
so class Child must be conforming to Codable by conformance to Parent,BUT properties of class Child won't be initialized from JSON properly.Decision I found is write all Codable stuff for class Child by myself, like:
class Child: Parent {
var token: String
var date : Date?
enum ChildKeys: CodingKey {
case token, date
}
required init(from decoder: Decoder) throws {
try super.init(from: decoder)
let container = try decoder.container(keyedBy: ChildKeys.self)
self.token = try container.decode(String.self, forKey: .token)
self.date = try container.decodeIfPresent(Date.self, forKey: .date)
}
override func encode(to encoder: Encoder) throws {
try super.encode(to: encoder)
var container = encoder.container(keyedBy: ChildKeys.self)
try container.encode(self.token, forKey: .token)
try container.encodeIfPresent(self.date, forKey: .date)
}
}
但是我觉得这是不对的,我错过了什么吗?如何在不编写所有内容的情况下使Child类正确地符合Codable?
But I feel it can't be right, did I missed something? How to make class Child conforming to Codable properly without writing all that stuff?
推荐答案
这是一篇不错的博客文章,其中包含您的问题的答案:来源
Here's a good blog post which includes an answer to your question: source
向下滚动到继承,您将看到以下内容:
假设我们有以下课程:
class Person : Codable {
var name: String?
}
class Employee : Person {
var employeeID: String?
}
我们通过继承Person类来获得Codable一致性,但是如果我们尝试对Employee的实例进行编码会发生什么呢?
We get the Codable conformance by inheriting from the Person class, but what happens if we try to encode an instance of Employee?
let employee = Employee()
employee.employeeID = "emp123"
employee.name = "Joe"
let encoder = JSONEncoder()
encoder.outputFormatting = .prettyPrinted
let data = try! encoder.encode(employee)
print(String(data: data, encoding: .utf8)!)
{
"name" : "Joe"
}
这不是预期的结果,因此我们必须添加这样的自定义实现:
This is not the expected result, so we have to add a custom implementation like this:
class Person : Codable {
var name: String?
private enum CodingKeys : String, CodingKey {
case name
}
func encode(to encoder: Encoder) throws {
var container = encoder.container(keyedBy: CodingKeys.self)
try container.encode(name, forKey: .name)
}
}
与子类相同的故事:
class Employee : Person {
var employeeID: String?
private enum CodingKeys : String, CodingKey {
case employeeID = "emp_id"
}
override func encode(to encoder: Encoder) throws {
var container = encoder.container(keyedBy: CodingKeys.self)
try container.encode(employeeID, forKey: .employeeID)
}
}
结果将是:
{
"emp_id" : "emp123"
}
这又不是预期结果,因此在这里,我们通过调用 super 使用继承力.
Which again is not the expected result, so here we are using inheritance by calling super.
// Employee.swift
override func encode(to encoder: Encoder) throws {
try super.encode(to: encoder)
var container = encoder.container(keyedBy: CodingKeys.self)
try container.encode(employeeID, forKey: .employeeID)
}
终于从一开始就给我们真正想要的东西:
Which finally gives us what we really wanted from the beginning:
{
"name": "Joe",
"emp_id": "emp123"
}
如果您对展平的结果不满意,那么也有避免这种情况的提示.
If you're not happy with the flattened result, there's a tip on how to avoid that too.
所有致谢撰写博客文章的人和我的感谢.希望它也对您有帮助,加油!
All the credits to the guy who wrote the blog post and my thanks.Hope it helps you as well, cheers!
这篇关于在Swift中默认使类符合Codable的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!