Json 如何从 Swift Codable 中排除属性

发布于06月21日

Swift's Encodable/Decodable protocols, released with Swift 4, make JSON (de)serialization quite pleasant. However, I have not yet found a way to have fine-grained control over which properties should be encoded and which should get decoded.

I have noticed that excluding the property from the accompanying CodingKeys enum excludes the property from the process altogether, but is there a way to have more fine-grained control?

推荐答案

要编码/解码的密钥列表由名为CodingKeys的类型控制(请注意末尾的s).编译器可以为您综合这一点，但始终可以覆盖它.

Let's say you want to exclude the property nickname from both encoding and decoding:

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(即编码而不是解码，反之亦然)，您必须提供您自己的encode(with encoder: )和init(from decoder: )实现:

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)
    }
}