CODEWITHSUNDEEP
scalacirce
martav
martav

Reputation: 1

How to Encode Map[K,V] using Circe

Reading at this documentation https://circe.github.io/circe/codecs/custom-codecs.html I understand that you should use the Circe KeyEncoder to define custom key types.

But I am confused if I want to encode a Map[K, V] where

case class K(a1: String, a2: String)
case class V(b1: String, b2: String)

Upvotes: 0

Views: 1189

Answers (1)

Taha
Taha

Reputation: 561

With class definitions;

case class K(a1: String, a2: String)
    
object K {
  import io.circe._
  implicit val encodeK: KeyEncoder[K] = (key: K) => s"${key.a1}-${key.a2}"
    
  implicit val decodeK: KeyDecoder[K] = (s: String) => {
    // Should be careful while parsing a K from String.
    val kParts = s.split("-")
    Some(K(kParts(0), kParts(1)))
  }
}

and

case class V(b1: String, b2: String)

object V {
  import io.circe._
  implicit val encodeV: Encoder[V] = (v: V) => Json.obj(
    ("b1", Json.fromString(v.b1)),
    ("b2", Json.fromString(v.b2))
  )

  implicit val decodeV: Decoder[V] = (c: HCursor) => for {
    b1 <- c.downField("b1").as[String]
    b2 <- c.downField("b2").as[String]
  } yield {
    new V(b1, b2)
  }
}

Conversion to/from JSON works as below:


object App {
  import io.circe.syntax._
  def main(args: Array[String]): Unit = {
    val map: Map[K, V] = Map(
      K("a11", "a21") -> V("b11", "b21"),
      K("a12", "a22") -> V("b12", "b22"),
      K("a13", "a23") -> V("b13", "b23")
    )
    
    val json = map.asJson
    println(json)
    println(json.as[Map[K, V]])
  }
}

Upvotes: 1

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