How to decode missing key as Option

[exoego]

Given

user1:
  name: John
  # user-type is missing
user2:
  name: Jane
  user-type: Admin

case class User(name: String, `user-type`: String) derives YamlDecoder

val decoded = yaml.as[Map[String, User]]

results in Left(ConstructError(Key user-type doesn't exist in parsed document))

I would like to decode missing key as Option.

[kpodsiad]

Hi @exoego, sorry for the long response time.

I'm afraid that currently user-type key ha to be present and only value can be omitted, see example with scala-cli:

//> using scala "3.1.1"
//> using lib "org.virtuslab::scala-yaml:0.0.4"
//> using lib "com.lihaoyi::pprint:0.7.3"

import org.virtuslab.yaml.*

object Yamltest extends App {

  case class User(name: String, `user-type`: Option[String]) derives YamlDecoder

  val yaml =
    s"""
        |user1:
        |  name: John
        |  user-type: 
        |user2:
        |  name: Jane
        |  user-type: Admin
        |""".stripMargin

  pprint.log(yaml.as[Map[String, User]])
}

Yaml.scala:21 yaml.as[Map[String, User]]: Right(
  value = Map(
    "user1" -> User(name = "John", user-type = None),
    "user2" -> User(name = "Jane", user-type = Some(value = "Admin"))
  )
)

[kpodsiad]

I agree that the library should be able to parse even missing key case.