Boilerplate free conversion from case classes to shapeless records via LabelledGeneric ... coming soon to shapeless 2.0.0.

gistfile1.scala

scala> import shapeless._, record._
import shapeless._
import record._

scala> case class Person(name: String, address: String, age: Int)
defined class Person

scala> val joe = Person("Joe", "Brighton", 33)
joe: Person = Person(Joe,Brighton,33)

scala> val gen = LabelledGeneric[Person]
gen: shapeless.LabelledGeneric[Person]{type Repr = shapeless.::[shapeless.record.FieldType[shapeless.tag.@@[Symbol,String("name")],String],shapeless.::[shapeless.record.FieldType[shapeless.tag.@@[Symbol,String("address")],String],shapeless.::[shapeless.record.FieldType[shapeless.tag.@@[Symbol,String("age")],Int],shapeless.HNil]]]} = $1$$1@19dbafd7

scala> val joeRecord = gen.to(joe)
joeRecord: gen.Repr = Joe :: Brighton :: 33 :: HNil

scala> joeRecord.keys
res0: shapeless.::[shapeless.tag.@@[Symbol,String("name")],shapeless.::[shapeless.tag.@@[Symbol,String("address")],shapeless.::[shapeless.tag.@@[Symbol,String("age")],shapeless.HNil]]] = 'name :: 'address :: 'age :: HNil

scala> joeRecord.values
res1: shapeless.::[String,shapeless.::[String,shapeless.::[Int,shapeless.HNil]]] = Joe :: Brighton :: 33 :: HNil

scala> joeRecord('name)
res2: String = Joe

scala> joeRecord('address)
res3: String = Brighton

scala> joeRecord('age)
res4: Int = 33

I would like to read the field name from a list. Such as;

val list = List('name, 'address, 'age)
list.map(x => lens[Person] >> x)

Is it possible?

Because I try to do this, and getting error. I also asked a similar question here. https://stackoverflow.com/questions/50418767/shapeless-lenses-usage-with-a-string-definition