Scoobie Scoobie - 1 year ago 73
Scala Question

How to deserialize without knowing concrete type using lift-json in scala?


2.0 and the following Scala classes & sealed trait:

sealed trait Location

case class Coordinate(latitude: Double,
longitude: Double) extends Location

case class Address(...) extends Location

I'd like to be able to deserialize a Location object without determining the concrete implementation:

deserialize[Location](""" { "latitude":0.0, "longitude":0.0 } """)

should produce the equivalent of:

val location: Location = Coordinate(0.0, 0.0)

Any way of doing this?

Answer Source

This may not be what you want, but with:

implicit val formats = net.liftweb.json.DefaultFormats
  .withHints(ShortTypeHints(List(classOf[Geo], classOf[Address])))

enables you to write

val loc: Location = read(write(Geo(0.0, 0.0)))

However your json then gets a TypeHint:


This formats can be played with somewhat, here is a nice post about type hints.

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