scala> val alnu = """foo ([a-zA-Z]+) ([0-9]+)""".r
alnu: scala.util.matching.Regex = foo ([a-zA-Z]+) ([0-9]+)

scala> val alnu (str, num) =  "foo bar 1234"
str: String = bar
num: String = 1234

works for me (2.8 final).

(a small hint on the website, how to format code, bold, cite, italic would be nice)

I don’t understand the syntax. Normally, in the REPL, if I write

val a = …
val a = …

the second a is hiding the first a – it’s a new declaration, and I may write

val a = 7
val a = “seven”

and the second line is, as if there never would have been the first line.

See:

http://www.scala-lang.org/node/122
http://fileit.in/p/7

object I {
def unapply(s: String): Option[Int] =
try { Some(s.toInt) }
catch { case _: NumberFormatException => None }
}

val LogEntry(I(totalTime), I(viewTime), I(dbTime), I(responseCode), uri) = line

totalTime: Int = 100
viewTime: Int = 25
dbTime: Int = 75
responseCode: Int = 200
uri: String =

object LogEntry {
def unapply(line: String) = {
val le(a,b,c,d,e) = line
Some(a.toInt, b.toInt, c.toInt, d, e)
}
}

Unfortunately, you can’t make Regex return Ints directly. Types are a compile-time thing, and your pattern can be any arbitrary String constructed at run-time, so in general there’s no way for the type-system to know what types of things you’ll be matching on. However, with a little compile-time help you can achieve what you want:

val le = “”"Completed in (\d+)ms \(View: (\d+), DB: (\d+)\) \| (\d+) OK \[http://app.domain.com(.*)\?.*""".r
val line = "Completed in 100ms (View: 25, DB: 75) | 200 OK [http://app.domain.com?params=here]”
object LogEntry {
def unapply(line: String): Option[(Int, Int, Int, String, String)] =
for (List(ttime, vtime, dtime, rc, uri) <- le.unapplySeq(line))
yield (ttime.toInt, vtime.toInt, dtime.toInt, rc, uri)
}
val LogEntry(totalTime, viewTime, dbTime, responseCode, uri) = line

totalTime: Int = 100
viewTime: Int = 25
dbTime: Int = 75
responseCode: String = 200
uri: String =

I am puzzling over returning integers though…