json - 如何从 Swift Codable 中排除属性？

Question

Swift 的 Encodable/Decodable 协议(protocol)与 Swift 4 一起发布，使 JSON(反)序列化非常愉快。但是，我还没有找到一种方法来细粒度地控制哪些属性应该被编码，哪些应该被解码。

我注意到从随附的 CodingKeys 枚举中排除该属性会完全将该属性排除在进程之外，但是有没有办法进行更细粒度的控制？

Answer 1

要编码/解码的键列表由称为 CodingKeys 的类型控制(注意末尾的 s)。编译器可以为您合成它，但始终可以覆盖它。

假设您想从编码和解码中排除属性nickname:

struct Person: Codable {
    var firstName: String
    var lastName: String
    var nickname: String?
    
    private enum CodingKeys: String, CodingKey {
        case firstName, lastName
    }
}

---

如果您希望它是非对称的(即编码但不解码，反之亦然)，您必须提供自己的 encode(with encoder: ) 和 init(from decoder :):

struct Person: Codable {
    var firstName: String
    var lastName: String
    
    // Since fullName is a computed property, it's excluded by default
    var fullName: String {
        return firstName + " " + lastName
    }

    private enum CodingKeys: String, CodingKey {
        case firstName, lastName, fullName
    }

    // We don't want to decode `fullName` from the JSON
    init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)
        firstName = try container.decode(String.self, forKey: .firstName)
        lastName = try container.decode(String.self, forKey: .lastName)
    }

    // But we want to store `fullName` in the JSON anyhow
    func encode(to encoder: Encoder) throws {
        var container = encoder.container(keyedBy: CodingKeys.self)
        try container.encode(firstName, forKey: .firstName)
        try container.encode(lastName, forKey: .lastName)
        try container.encode(fullName, forKey: .fullName)
    }
}