Scala: convert map to case class

Question

Let\'s say I have this example case class

case class Test(key1: Int, key2: String, key3: String)

And I have a map

myMap = M         


        

Answer 1

Here is an alternative non-boilerplate method that uses Scala reflection (Scala 2.10 and above) and doesn't require a separately compiled module:

import org.specs2.mutable.Specification
import scala.reflect._
import scala.reflect.runtime.universe._

case class Test(t: String, ot: Option[String])

package object ccFromMap {
  def fromMap[T: TypeTag: ClassTag](m: Map[String,_]) = {
    val rm = runtimeMirror(classTag[T].runtimeClass.getClassLoader)
    val classTest = typeOf[T].typeSymbol.asClass
    val classMirror = rm.reflectClass(classTest)
    val constructor = typeOf[T].decl(termNames.CONSTRUCTOR).asMethod
    val constructorMirror = classMirror.reflectConstructor(constructor)

    val constructorArgs = constructor.paramLists.flatten.map( (param: Symbol) => {
      val paramName = param.name.toString
      if(param.typeSignature <:< typeOf[Option[Any]])
        m.get(paramName)
      else
        m.get(paramName).getOrElse(throw new IllegalArgumentException("Map is missing required parameter named " + paramName))
    })

    constructorMirror(constructorArgs:_*).asInstanceOf[T]
  }
}

class CaseClassFromMapSpec extends Specification {
  "case class" should {
    "be constructable from a Map" in {
      import ccFromMap._
      fromMap[Test](Map("t" -> "test", "ot" -> "test2")) === Test("test", Some("test2"))
      fromMap[Test](Map("t" -> "test")) === Test("test", None)
    }
  }
}