Groovy syntax for regular expression matching
RegexGroovyRegex 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"
}