json parsing using circe in scala

Question

I'm trying to make use of circe for json parsing in scala. Can you please help me parse 'pocs' from the data in the case class as well? here is the code:

import io.circe.Decoder
import io.circe.generic.semiauto.deriveDecoder
import io.circe.parser

val json: String =
"""
{
    "segmements": [
        {
            "tableName": "X",
            "segmentName": "XX",
            "pocs": [
                "[email protected]",
                "[email protected]"
            ]
        },
        {
            "tableName": "Y",
            "segmentName": "YY",
            "pocs": [
                "[email protected]",
                "[email protected]"
            ]
        }
    ]
}
"""

final case class TableInfo(tableName: String, segmentName: String)
object TableInfo {
  implicit final val TableInfoDecoder: Decoder[TableInfo] = deriveDecoder
}

val result = for {
  data <- parser.parse(json)
  obj <- data.asObject.toRight(left = new Exception("Data was not an object"))
  segmements <- obj("segmements").toRight(left = new Exception("Json didn't had the 
segments key"))
  r <- segmements.as[List[TableInfo]]
} yield r

println(result)

scastie link: https://scastie.scala-lang.org/BalmungSan/eVEvBulOQwGzg5hIJroAoQ/3

Guru Stron · Accepted Answer · 2022-08-04 17:26:07Z

1

Just add parameter typed as collection of String:

final case class TableInfo(tableName: String, segmentName: String, pocs: Seq[String])

scastie

answered Aug 4, 2022 at 17:26

Guru Stron

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2 Comments

Vish0111 Over a year ago

Thank you for the answer. I'm a little new to Scala, Can you please tell me a little more about the last line of code Right(List(TableInfo( I couldn't understand it clearly. Thank you.

Guru Stron Over a year ago

@Vish0111 it's scastie feature - basically it outputs the value of result (like println but without it) with type info. You can just remove result and it will disappear also You can see the same near println - (): Unit which basically return of println.

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