CODEWITHSUNDEEP
regexraku
Zarul Zakuan
Zarul Zakuan

Reputation: 520

Perl6 assign regex match groups to variables

In P5, I am able to do something like this

my ($var1, $var2, $var3) = $string =~ /(.+)\s(.+)\s(.+)/;

How do I do the same in Perl 6? If I do the same syntax, the $var1 will hold the entire $string value.

Upvotes: 6

Views: 214

Answers (2)

timotimo
timotimo

Reputation: 4329

The result from that match is a Match object, which by itself won't behave like a list, and therefore won't expand into the three variables. However, the Match object has a "list" method that does what you want. Here's an example:

my $string = "hello how are you";
my ($var1, $var2, $var3) =
    ($string ~~ /(.+)\s(.+)\s(.+)/).list;
say "$var1 and $var2 and $var3
# Output: hello how and are and you

A few further things to point out:

  • Since .+ is a greedy match, and it also accepts spaces, the first capture will eat up two words.
  • Surely the code in the question is a simplified example, but if you ever need to split text into words, maybe the words method does what you want. Of course, you'll want to check what exactly you want: Split by spaces? Only return alphabetic characters, so that periods and commas aren't in the final result? etc.
  • If you do need to match the same thing multiple times, maybe the comb method is actually more helpful?

Upvotes: 8

Tomalak
Tomalak

Reputation: 338148

my $string = 'foo bar baz';

my $match = $string ~~ /(.+)\s(.+)\s(.+)/;
say $match;     # 'foo bar baz'
say $match[0];  # 'foo'
say $match[1];  # 'bar'
say $match[2];  # 'baz'

my ($foo, $bar, $baz) = @$match;
say $foo;       # 'foo'
say $bar;       # 'bar'
say $baz;       # 'baz'

therefore

my ($foo, $bar, $baz) = @($string ~~ /(.+)\s(.+)\s(.+)/);

Upvotes: 7

Related Questions