Hello superwombat,

If you precede the expression to match with the greedy match-all .*, the match you get will be the last one:

#! perl
use strict;
use warnings;

my $string = '
start:
end:

test
code
start: real
1
end: real
with
start: real
repeating newlines
and more than
start: real
one

instance
end: real
of the 
start: real
desired
string


end: real


start:
end:';

my $header = 'start: real';
my $footer = 'end: real';

$string =~ /.*(^$header$(?:.*)^$footer$)/ms;

print "$1\n" if $1;

[download]

Output:

18:45 >perl 1078_SoPW.pl
start: real
desired
string


end: real

18:45 >

[download]

Note the /ms modifiers in the regex. As perlre explains:

Used together, as /ms, they let the "." match any character whatsoever, while still allowing "^" and "$" to match, respectively, just after and just before newlines within the string.

Hope that helps,

Athanasius <°(((>< contra mundum Iustus alius egestas vitae, eros Piratica,

In reply to Re: Regex to find the last matching set in a long scalar by Athanasius
in thread Regex to find the last matching set in a long scalar by superwombat

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