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Ignore non-Codable optional properties in Codable object

When conforming to Codable protocol, I cannot easily skip optional property of non-Codable class

In Ride struct we want to skip encoding and decoding of driver property, and just leave it nil when decoding:

  struct Ride: Codable {

    public var number: String
    public var passenger: Passenger? // Codable conforming
    public var driver: Driver?       // NSObject subclass, doesn't conform to Codable

    enum CodingKeys: String, CodingKey {
      case number
      case passenger
    }
  }

In this case I get a compile error

Type 'Driver' does not conform to protocol 'Decodable'

So far the only solution I came with, is to encode and decode manually by providing following methods:

public init(from decoder: Decoder) throws {
    let values = try decoder.container(keyedBy: CodingKeys.self)
    number = try? values.decode(String.self, forKey: .number)
    passenger = try? values.decode(Passenger.self, forKey: .passenger)
}

public func encode(to encoder: Encoder) throws {
    var container = encoder.container(keyedBy: CodingKeys.self)
    try container.encode(number, forKey: .number)
    try container.encode(passenger, forKey: .passenger)
}

Is there another way to skip an optional property? This would be much easier with bigger models.

like image 799
KIO Avatar asked Oct 18 '17 15:10

KIO


1 Answers

Try providing a default value for driver. For instance, this compiled without errors on Xcode 9.0:

struct Ride: Codable {
    public var number: String
    public var passenger: Passenger? 
    public var driver: Driver? = nil

    private enum CodingKeys: String, CodingKey {
        case number
        case passenger
    }
}

struct Passenger: Codable { /* ... */ }

class Driver: NSObject { /* ... */ }

Quick test:

let rideJSON =
"""
{
    "number": "123"
}
""".data(using: .utf8)!

let ride = try! JSONDecoder().decode(Ride.self, from: rideJSON)
print(ride) /* Ride(number: "123", passenger: nil, driver: nil) */

By the way, I used a private CodingKeys because that is what the compiler does by default as well ;)

like image 61
Paulo Mattos Avatar answered Sep 19 '22 14:09

Paulo Mattos