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.
What is Group in Regex? A group is a part of a regex pattern enclosed in parentheses () metacharacter. We create a group by placing the regex pattern inside the set of parentheses ( and ) . For example, the regular expression (cat) creates a single group containing the letters 'c', 'a', and 't'.
$1 equals the text " brown ".
Simple word matching In this statement, World is a regex and the // enclosing /World/ tells Perl to search a string for a match. The operator =~ associates the string with the regex match and produces a true value if the regex matched, or false if the regex did not match.
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:
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.comb
method is actually more helpful?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(.+)/);
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