在Swift中默认使类符合Codable [英] Make class conforming to Codable by default in Swift

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": "[e-mail]", "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:

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!

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