CODEWITHSUNDEEP
regexperlcapture
Aaron Moodie
Aaron Moodie

Reputation: 1965

How can I capture only one part of Perl regular expression with grouping?

I've currenly trying to pull out dates from a file and feed them directly into an array. My regex is working, but I have 6 groups in it, all of which are being added to the array, when I only want the first one.

@dates = (@dates, ($line =~ /((0[1-9]|[12][0-9]|3[01])(\/|\-)(0[1-9]|1[0-2])(\/|\-)([0-9][0-9][0-9][0-9]|[0-9][0-9]))/g ));

is there a simple way to grab the $1 group of a perl regex?

my output is looking like this:

13/04/2009, 13, /, 04, /, 2009, 14-12-09, 14, -, 12, -, 09

Upvotes: 1

Views: 337

Answers (2)

justintime
justintime

Reputation: 3631

That regex looks like the sort of thing that would confuse me when I next pick up the code. I would break it out :

    my $date= qr/
       (?:0[1-9]|[12][0-9]|3[01])           # day
       (?:\/|\-)
       (?:0[1-9]|1[0-2])                    # month
       (?:\/|\-)
       (?:[0-9][0-9][0-9][0-9]|[0-9][0-9])  #year
       /x ;

You can add the element onto the array using 

    push @dates,   ($line =~ /($date)/ ) ;

You an simplify the seperator bit (notice I have changed to using ( ) rather than / / to avoid having to backslahs the /

    my $date= qr (
       (?: 0[1-9] | [12][0-9] | 3[01]       # day
       [/-]
       (?:0[1-9]|1[0-2])                    # month
       [/-]
       (?:\d{4}|\d{2})                      #year
       )x ;

Upvotes: 4

Aaron Moodie
Aaron Moodie

Reputation: 1965

just found it. You can create a passive group by using ?: at the start of the group.

@dates = (@dates, ($line =~ /((?:0[1-9]|[12][0-9]|3[01])(?:\/|\-)(?:0[1-9]|1[0-2])(?:\/|\-)(?:[0-9][0-9][0-9][0-9]|[0-9][0-9]))/g ));

by making all other groups passive, now only the first group is added to the array.

Upvotes: 1

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