How to decode a JSON property with different types? [duplicate]

Answer 1

You will have to implement your own func encode(to encoder: Encoder) throws and init(from decoder: Decoder) throws which are both properties of the Codable protocol. Then change your rating variable into an enum

Which would look like this:

enum Rating: Codable {
    case int(Int)
    case string(String)

    func encode(to encoder: Encoder) throws {
        var container = encoder.singleValueContainer()
        switch self {
        case .int(let v): try container.encode(v)
        case .string(let v): try container.encode(v)
        }
    }

    init(from decoder: Decoder) throws {
        let value = try decoder.singleValueContainer()

        if let v = try? value.decode(Int.self) {
            self = .int(v)
            return
        } else if let v = try? value.decode(String.self) {
            self = .string(v)
            return
        }

        throw Rating.ParseError.notRecognizedType(value)
    }

    enum ParseError: Error {
        case notRecognizedType(Any)
    }
}

Then on your DetailModel just change rating: String to rating: Rating

This works, I have tested with these JSON strings.

// int rating
{   
    "rating": 200,
    "bro": "Success"
}

// string rating
{
    "rating": "200",
    "bro": "Success"
}

Edit: I've found a better swiftier way of implementing init(from decoder: Decoder) throws, which produces a better error message, by using this you can now omit the ParseError enum.

init(from decoder: Decoder) throws {
    let value = try decoder.singleValueContainer()
    do {
        self = .int(try value.decode(Int.self))
    } catch DecodingError.typeMismatch {
        self = .string(try value.decode(String.self))
    }
}