You're assuming the first occurrence of the "b" is after the first occurrence of the "a". Watch it break — after the fix as done by davis: 
my $str = 'xbaaaaaa' . ('_' x 20) . 'bcccccccccc';
my $start = $str =~ /a+/ ? $+[0] : 0;
my $end = $str =~ /b/ ? $-[0] : 0;

my $ext = substr $str, $start, $end - $start;
$, = " | "; $\ = "\n";
print $start, $end, $ext;

[download]
Which prints: 
8 | 1 | ____________________bccc

You need to continue the second search where the first one left off. Adding the //g switch to both regexps can do that, provided you make sure pos is clear before you start on the first one. A failed match can take care of that. I'm not sure that's absolutely necessary, but you never know... It depends of what you matched on $str before, and on whether pos gets properly localised to the current block, by perl. (Note: it doesn't make a difference if you do it or not for this particular string, but I'm trying to cover all possibilities, in general. I want to make sure the first regexp always starts searching from the start of the string.) 

my $str = 'xbaaaaaa' . ('_' x 20) . 'bcccccccccc';
$str =~ /(?!)/g;   # A match that always fails, resetting pos()
my $start = $str =~ /a+/g ? $+[0] : 0;
my $end = $str =~ /b/g ? $-[0] : 0;

my $ext = substr $str, $start, $end - $start;
$, = " | "; $\ = "\n";
print $start, $end, $ext;

[download]
resulting in: 
8 | 28 | ____________________

In reply to Re: Re: print data between two regular expressions by bart
in thread print data between two regular expressions by Anonymous Monk

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