In Scala, how do I get the *name* of an `object` (

问题:

In Scala, I can declare an object like so:

class Thing

object Thingy extends Thing

How would I get "Thingy" (the name of the object) in Scala?

I've heard that Lift (the web framework for Scala) is capable of this.

回答1:

Just get the class object and then its name.

scala> Thingy.getClass.getName
res1: java.lang.String = Thingy$

All that's left is to remove the $.

EDIT:

To remove names of enclosing objects and the tailing $ it is sufficient to do

res1.split("\\$").last

回答2:

If you declare it as a case object rather than just an object then it'll automatically extend the Product trait and you can call the productPrefix method to get the object's name:

scala> case object Thingy
defined module Thingy

scala> Thingy.productPrefix
res4: java.lang.String = Thingy

回答3:

I don't know which way is the proper way, but this could be achieved by Scala reflection:

implicitly[TypeTag[Thingy.type]].tpe.termSymbol.name.toString