Scala capture group using regex

Question

Let's say I have this code:

val string = "one493two483three"
val pattern = """two(\d+)three""".r
pattern.findAllIn(string).foreach(println)

I expected findAllIn to only return 483, but instead, it returned two483three. I know I could use unapply to extract only that part, but I'd have to have a pattern for the entire string, something like:

 val pattern = """one.*two(\d+)three""".r
 val pattern(aMatch) = string
 println(aMatch) // prints 483

Is there another way of achieving this, without using the classes from java.util directly, and without using unapply?

Oversweet

Answer 1

Here's an example of how you can access group(1) of each match:

val string = "one493two483three"
val pattern = """two(\d+)three""".r
pattern.findAllIn(string).matchData foreach {
   m => println(m.group(1))
}

This prints "483" (as seen on ideone.com).

---

The lookaround option

Depending on the complexity of the pattern, you can also use lookarounds to only match the portion you want. It'll look something like this:

val string = "one493two483three"
val pattern = """(?<=two)\d+(?=three)""".r
pattern.findAllIn(string).foreach(println)

The above also prints "483" (as seen on ideone.com).

References

• regular-expressions.info/Lookarounds

Nunley

You can also use pattern.findAllMatchIn(string).foreach... instead – Hardi

Answer 2

val string = "one493two483three"
val pattern = """.*two(\d+)three.*""".r

string match {
  case pattern(a483) => println(a483) //matched group(1) assigned to variable a483
  case _ => // no match
}

Heriot

This is the simplest way by far. You use the regex object ("pattern") in a match/case and extracts the group into the variable a483. The problem withthis case is that the pattern should have wildcards on both sides: val pattern = """.*two(\d+)three.*""".r – Cranmer

Yes. I don't think the above is immediately clear, but once you understand that it's assigning the digit matching group to the variable 'a483', then it makes more sense. Perhaps rewrite in a clearer fashion ? – Trixi

This is the scala way with regex. For people don't understand the magic behind this answer, try search "scala regex extractor" or "scala unapply regex" etc. – Haploid

the semantics is unclear. is this the first, last, or a random match from the string? – Rescind

Answer 3

Starting Scala 2.13, as an alternative to regex solutions, it's also possible to pattern match a String by unapplying a string interpolator:

"one493two483three" match { case s"${x}two${y}three" => y }
// String = "483"

Or even:

val s"${x}two${y}three" = "one493two483three"
// x: String = one493
// y: String = 483

If you expect non matching input, you can add a default pattern guard:

"one493deux483three" match {
  case s"${x}two${y}three" => y
  case _                   => "no match"
}
// String = "no match"

Chema

Answer 4

You want to look at group(1), you're currently looking at group(0), which is "the entire matched string".

See this regex tutorial.

Repine

can you illustrate on the input I provided? I tried to call group(1) on what's returned by findAllIn but I get an IllegalStateException. – Oversweet

Answer 5

def extractFileNameFromHttpFilePathExpression(expr: String) = {
//define regex
val regex = "http4.*\\/(\\w+.(xlsx|xls|zip))$".r
// findFirstMatchIn/findAllMatchIn returns Option[Match] and Match has methods to access capture groups.
regex.findFirstMatchIn(expr) match {
  case Some(i) => i.group(1)
  case None => "regex_error"
}
}
extractFileNameFromHttpFilePathExpression(
    "http4://testing.bbmkl.com/document/sth1234.zip")

Steamtight