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storing match into other scalar

zerysaish
New User

Apr 7, 2010, 8:32 AM

Post #1 of 5 (1928 views)

storing match into other scalar	Can't Post

i have $string = "We'll meet again on 20/10/2010";
$string2 = "10/10/2012 sounds better";

how can i extract those date using regex, and store them into other scalar

eg. $date1 = 20/10/2010
$date2 = 10/10/2012

roolic
User

Apr 7, 2010, 8:48 PM

Post #2 of 5 (1916 views)

Re: [zerysaish] storing match into other scalar [In reply to]	Can't Post

Code

# 1 
my $date; 
if( $string =~/(\d{2}\/\d{2}\/\d{4})/ ){ $date = $1; } 
# 2 
my ($date) = ($string =~/(\d{2}\/\d{2}\/\d{4})/g ); # first 
my @dates = ($string =~/(\d{2}\/\d{2}\/\d{4})/g ); # all

zerysaish
New User

Apr 8, 2010, 8:48 AM

Post #3 of 5 (1906 views)

Re: [roolic] storing match into other scalar [In reply to]	Can't Post

if($string =~ m{\b(
\d{1,2}
/
\d{1,2}
/
/d{2,4})
\b}gx)
{
@dates = ????
}

how can u store all the matches found into that @dates array? if let say $string = "01/12/1990 11/11/2000 12/31/2000"

rovf
Veteran

Apr 9, 2010, 12:52 AM

Post #4 of 5 (1899 views)

Re: [zerysaish] storing match into other scalar [In reply to]	Can't Post

In your regexp, you use grouping (parentheses; see the perlre man-page) to capture the patterns you are interested in. Then you have two possibilities:

(1) If you evaluate the pattern match in list context, i.e.

@s=$x =~ /...(....)....(....)..../

the result contains the parts of the string which match the groups. This is discussed in the perlop man-page (scroll
forward to the section "Regexp Quote-Like Operators").

(2) You can pick up $1, $2 etc. after the match.

zerysaish
New User

Apr 10, 2010, 9:39 AM

Post #5 of 5 (1857 views)

Re: [rovf] storing match into other scalar [In reply to]	Can't Post

thank u very much for ur assistance, i managed to find out how to do that.

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