Unapply regex matching into case class

Question

Given a regex and a string

val reg = "(a)(b)"
val str = "ab"

and a corresponding case class

case class Foo(a: string, b: string)

How can I match the regex against the string and unapply the matches into the case class so I have

Foo("a", "b")

in the end?

Answer 1

Pattern match on the result of finding the regular expression in the string and assigning the results to Foo . The API docs for Regex have a similar example.

scala> val reg = "(a)(b)".r
reg: scala.util.matching.Regex = (a)(b)

scala> val str = "ab"
str: String = ab

scala> case class Foo(a: String, b: String)
defined class Foo

scala> val foo = reg.findFirstIn(str) match{
     | case Some(reg(a,b)) => new Foo(a,b)
     | case None => Foo("","")
     | }
foo: Foo = Foo(a,b)

 scala> foo.a
res2: String = a

scala> foo.b
res3: String = b

Answer 2

If you are sure that the match is ok, you can currify the function and use a foldleft to apply this version to the list of extracted arguments:

> val reg = "(a)(b)".r
> val str = "ab"
> val f1 = (Foo.apply _).curried
// f1 : String => String => Foo
> val groups = reg.findFirstMatchIn(str).get.subgroups
// groups: List[String]
> val foo = ((f1 : Any) /: groups) { case (f: (String => Any), s) => f(s)} asInstanceOf[Foo]

foo: Foo = Foo(a,b)

See Function2 in scala for the definition of curried.