Groovy syntax for regular expression matching

RegexGroovy

Regex Problem Overview


What is the Groovy equivalent of the following Perl code?

my $txt = "abc : groovy : def";
if ($txt =~ / : (.+?) : /) {
  my $match = $1;
  print "MATCH=$match\n"; 
  # should print "MATCH=groovy\n"
}

I know that there's more than one way to do it (including the regular Java way) - but what is the "Groovy way" of doing it?

This is one way of doing it, but it feels a bit clumsy - especially the array notation (m[0][1]) which feels a bit strange. Is there a better way do it? If not - please describe the logic behind m[0][1].

def txt = "java : groovy : grails"
if ((m = txt =~ / : (.+?) :/)) {
  def match = m[0][1]
  println "MATCH=$match"
}

Regex Solutions


Solution 1 - Regex

m[0] is the first match object.
m[0][0] is everything that matched in this match.
m[0][1] is the first capture in this match.
m[0][2] is the second capture in this match.

Based on what I have read (I don't program in Groovy or have a copy handy), given

def m = "barbaz" =~ /(ba)([rz])/;

m[0][0] will be "bar"
m[0][1] will be "ba"
m[0][2] will be "r"
m[1][0] will be "baz"
m[1][1] will be "ba"
m[1][2] will be "z"

I could not stand not knowing if I was correct or not, so I downloaded groovy and wrote an example:

def m = "barbaz" =~ /(ba)([rz])/;

println "m[0][0] " + m[0][0]
println "m[0][1] " + m[0][1]
println "m[0][2] " + m[0][2]
println "m[1][0] " + m[1][0]
println "m[1][1] " + m[1][1]
println "m[1][2] " + m[1][2]

Solution 2 - Regex

This was the closest match to the Perl code that I could achieve:

def txt = "abc : groovy : def"
if ((m = txt =~ / : (.+?) : /)) {
  def match = m.group(1)
  println "MATCH=$match"
}

// Prints:
// MATCH=groovy

Solution 3 - Regex

This is my best understanding of how to do this using Groovy syntax (but see lfaraone's response too):

import java.util.regex.Matcher

def txt = 'abc : groovy : def'
if (txt =~ ~/ : (.+?) : /) {
    def match = Matcher.lastMatcher[0][1]
    println "MATCH=$match"
}

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionknorvView Question on Stackoverflow
Solution 1 - RegexChas. OwensView Answer on Stackoverflow
Solution 2 - RegexknorvView Answer on Stackoverflow
Solution 3 - RegexChris Jester-YoungView Answer on Stackoverflow