Exclude CodingKeys that doesn't need to be altered?

Question

Say I have a struct User model which has many properties in it.

struct User: Codable {
    let firstName: String
    let lastName: String
    // many more properties...
}

As you can see above it conforms to Codable. Imagine if the lastName property is should be encoded/decoded as secondName and I would like to keep it as lastName at my end, I need to add the CodingKeys to the User model.

struct User: Codable {
    //...
    private enum CodingKeys: String, CodingKey {
        case firstName
        case lastName = "secondName"
        // all the other cases...
    }
}

Is there any possible way to avoid including all the cases in CodingKeys that have the same value as rawValue like the firstName in the above example (Feels redundant)? I know if I avoid the cases in CodingKeys it won't be included while decoding/encoding. But, is there a way I could override this behaviour?

Answer 1

There is a codable way, but the benefit is questionable.

Create a generic CodingKey

struct AnyKey: CodingKey {
    var stringValue: String
    var intValue: Int?

    init?(stringValue: String) { self.stringValue = stringValue; self.intValue = nil }
    init?(intValue: Int) { self.stringValue = String(intValue); self.intValue = intValue }
}

and add a custom keyDecodingStrategy

struct User: Codable {
    let firstName: String
    let lastName: String
    let age : Int
}

let jsonString = """
{"firstName":"John", "secondName":"Doe", "age": 30}
"""

let data = Data(jsonString.utf8)

do {
    let decoder = JSONDecoder()
    decoder.keyDecodingStrategy = .custom({ keyPath -> CodingKey in
        let key = keyPath.last!
        return key.stringValue == "secondName" ? AnyKey(stringValue:"lastName")! : key
    })
    let result = try decoder.decode(User.self, from: data)
    print(result)
} catch {
    print(error)
}