Case class to map in Scala

Question

Is there a nice way I can convert a Scala case class instance, e.g.

case class MyClass(param1: String, param2: String)
val x = MyClass("hello", "world")

into a mapping of some kind, e.g.

getCCParams(x) returns "param1" -> "hello", "param2" -> "world"

Which works for any case class, not just predefined ones. I've found you can pull the case class name out by writing a method that interrogates the underlying Product class, e.g.

def getCCName(caseobj: Product) = caseobj.productPrefix 
getCCName(x) returns "MyClass"

So I'm looking for a similar solution but for the case class fields. I'd imagine a solution might have to use Java reflection, but I'd hate to write something that might break in a future release of Scala if the underlying implementation of case classes changes.

Currently I'm working on a Scala server and defining the protocol and all its messages and exceptions using case classes, as they are such a beautiful, concise construct for this. But I then need to translate them into a Java map to send over the messaging layer for any client implementation to use. My current implementation just defines a translation for each case class separately, but it would be nice to find a generalised solution.

Answer 1

If you happen to be using Json4s, you could do the following:

import org.json4s.{Extraction, _}

case class MyClass(param1: String, param2: String)
val x = MyClass("hello", "world")

Extraction.decompose(x)(DefaultFormats).values.asInstanceOf[Map[String,String]]

Answer 2

Here's a simple variation if you don't care about making it a generic function:

case class Person(name:String, age:Int)

def personToMap(person: Person): Map[String, Any] = {
  val fieldNames = person.getClass.getDeclaredFields.map(_.getName)
  val vals = Person.unapply(person).get.productIterator.toSeq
  fieldNames.zip(vals).toMap
}

scala> println(personToMap(Person("Tom", 50)))
res02: scala.collection.immutable.Map[String,Any] = Map(name -> Tom, age -> 50)

Answer 3

I don't know about nice... but this seems to work, at least for this very very basic example. It probably needs some work but might be enough to get you started? Basically it filters out all "known" methods from a case class (or any other class :/ )

object CaseMappingTest {
  case class MyCase(a: String, b: Int)

  def caseClassToMap(obj: AnyRef) = {
    val c = obj.getClass
    val predefined = List("$tag", "productArity", "productPrefix", "hashCode",
                          "toString")
    val casemethods = c.getMethods.toList.filter{
      n =>
        (n.getParameterTypes.size == 0) &&
        (n.getDeclaringClass == c) &&
        (! predefined.exists(_ == n.getName))

    }
    val values = casemethods.map(_.invoke(obj, null))
    casemethods.map(_.getName).zip(values).foldLeft(Map[String, Any]())(_+_)
  }

  def main(args: Array[String]) {
    println(caseClassToMap(MyCase("foo", 1)))
    // prints: Map(a -> foo, b -> 1)
  }
}

Answer 4

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

By zipping field names with field values obtained with productIterator we can generically obtain the associated Map:

// case class MyClass(param1: String, param2: String)
// val x = MyClass("hello", "world")
(x.productElementNames zip x.productIterator).toMap
// Map[String,Any] = Map("param1" -> "hello", "param2" -> "world")

Answer 5

commons.mapper.Mappers.Mappers.beanToMap(caseClassBean)

Details: https://github.com/hank-whu/common4s

Answer 6

This should work:

def getCCParams(cc: AnyRef) =
  (Map[String, Any]() /: cc.getClass.getDeclaredFields) {(a, f) =>
    f.setAccessible(true)
    a + (f.getName -> f.get(cc))
  }