Obtaining the full info about fields of the case class in Scala

Question

Consider the following class and method:

case class User(id: Long, name: String) {
  private var foo = "Foo" // shouldn't be printed
  val bar = "bar" // also shouldn't be printed
}
case class Message(id: Long, userId: Long, text: String)

def printInfo[E](o: E)(implicit tt: TypeTag[E]) = {

}

I want to make this method print the name, type and value for each of the field for any case class, i.e.

printInfo(User(1, "usr1")) // prints something like "(id, Long, 1), (name, String)"
printInfo(Message(1, 1, "Hello World")) // prints "(id, Long, 1), (userId, Long, 1), (text, String, "Hello World")"

Adding some custom annotations for fields is also considerable.

This is almost a duplicate of your [previous question](http://stackoverflow.com/questions/42189259/obtaining-class-name-and-property-names-from-generic-class). — Jasper-M, Feb 15 '17 at 15:16

score 19 · Accepted Answer · answered Feb 15 '17 at 13:35

You could do this by inspecting the members listed by the type tag and reflecting using its mirror:

import scala.reflect.ClassTag
import scala.reflect.runtime.universe.TypeTag

def printInfo[A](a: A)(implicit tt: TypeTag[A], ct: ClassTag[A]): String = {
  val members = tt.tpe.members.collect {
    case m if m.isMethod && m.asMethod.isCaseAccessor => m.asMethod
  }

  members.map { member =>
    val memberValue = tt.mirror.reflect(a).reflectMethod(member)()
    s"(${ member.name }, ${ member.returnType }, $memberValue)"
  }.mkString(", ")
}

Which would work like this:

scala> case class User(id: Long, name: String) {
     |   private var foo = "Foo" // shouldn't be printed
     |   val bar = "bar" // also shouldn't be printed
     | }
defined class User

scala> case class Message(id: Long, userId: Long, text: String)
defined class Message

scala> printInfo(User(1, "usr1"))
res0: String = (name, String, usr1), (id, scala.Long, 1)

scala> printInfo(Message(1, 1, "Hello World"))
res1: String = (text, String, Hello World), (userId, scala.Long, 1), (id, scala.Long, 1)

(If you wanted Long instead of scala.Long it wouldn't be too hard to drop the prefix from the type you get from member.returnType, but I'll leave that as an exercise for the reader.)

It's also not too hard to do this without any runtime reflection using Shapeless:

import shapeless.{ ::, HList, HNil, LabelledGeneric, Typeable, Witness }
import shapeless.labelled.FieldType

trait PrettyPrintable[A] {
  def apply(a: A): List[(String, String, String)]
}

object PrettyPrintable {
  implicit val hnilPrettyPrintable: PrettyPrintable[HNil] =
    new PrettyPrintable[HNil] {
      def apply(a: HNil): List[(String, String, String)] = Nil
    }

  implicit def hconsPrettyPrintable[K <: Symbol, H, T <: HList](implicit
    kw: Witness.Aux[K],
    ht: Typeable[H],
    tp: PrettyPrintable[T]
  ): PrettyPrintable[FieldType[K, H] :: T] =
    new PrettyPrintable[FieldType[K, H] :: T] {
      def apply(a: FieldType[K, H] :: T): List[(String, String, String)] =
        (kw.value.name, ht.describe, a.head.toString) :: tp(a.tail)
    }

  implicit def genPrettyPrintable[A, R <: HList](implicit
    ag: LabelledGeneric.Aux[A, R],
    rp: PrettyPrintable[R]
  ): PrettyPrintable[A] = new PrettyPrintable[A] {
    def apply(a: A): List[(String, String, String)] = rp(ag.to(a))
  }

  def printInfo[A](a: A)(implicit pp: PrettyPrintable[A]) = pp(a).map {
    case (memberName, memberType, memberValue) =>
      s"($memberName, $memberType, $memberValue)"
  }.mkString(", ")
}

And then:

scala> PrettyPrintable.printInfo(User(1, "usr1"))
res2: String = (id, Long, 1), (name, String, usr1)

scala> PrettyPrintable.printInfo(Message(1, 1, "Hello World"))
res3: String = (id, Long, 1), (userId, Long, 1), (text, String, Hello World)

Among other things this gives you the fields in declaration order, which I think should be possible with the type tag approach, but I avoid that API as often as I can, so off the top of my head I'm not sure.

score 3 · Answer 2 · answered Oct 09 '18 at 19:51

Starting Scala 2.13, case classes (which are an implementation of Product) are now provided with a productElementNames method which returns an iterator over their field's names.

Combined with productIterator:

// val user = User(1, "user")
(user.productElementNames zip user.productIterator)
  .map { case (field, value) => (field, value.getClass.getSimpleName, value) }
  .toList
// List[(String, String, Any)] = List((id,Long,1), (name,String,user))

This:

extracts field names (Iterator(id, name)) using productElementNames
extracts field values (Iterator(1, user)) using productIterator
zips field names with field values
extracts the class name of values

Obtaining the full info about fields of the case class in Scala

2 Answers2