CODEWITHSUNDEEP
scala
akauppi
akauppi

Reputation: 18036

Initializing Scala's Option[String] from a string so that certain value leads to None

Is there a shorter way to doing this:

val accessControlAllowOrigin = c_dash.getString("access-control-allow-origin") match {
  case "" => None
  case x => Some(x)
}

This is reading in a Typesafe Config value, where empty string denotes the absence of such a config (in Typesafe Config it's good manners to include all values, not leaving anything out).

Is there something like:

val sopt = Option( s, "magic" )

..which would provide either Some(s) or None if s's value is "magic"?

By looking at the doc I came to:

scala> def f(s: String) = (Some(s) filter( _ != "magic" ))
f: (s: String)Option[String]

scala> f("aaa")
res1: Option[String] = Some(aaa)

scala> f("magic")
res2: Option[String] = None

Is that the simplest?

Upvotes: 1

Views: 426

Answers (2)

The Archetypal Paul
The Archetypal Paul

Reputation: 41749

I may be missing something obvious, but how about:

val sopt = if (s == "magic") None else Some(s)

Or, for your f():

def f(s:String) = if (s == "magic") None else Some(s)

or, more generically:

def noneIfDefault(s:String, default: String) = if (s == default) None else Some(s)

Upvotes: 2

Michael Zajac
Michael Zajac

Reputation: 55569

You can use filter, which works the same for an Option as it would for any other collection:

c_dash.getString("access-control-allow-origin").filter(_.nonEmpty)

Anything not matching the filter predicate will become None.

Upvotes: 5

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