One point, s/// isn't "sed" ... it's "substitute".

Perl, of course, got its s/// from sed...

Of course. And even in sed, s/// means "substitute".

-sauoq
"My two cents aren't worth a dime.";

Full code. I am attempting to get a list of the ttys associated with the running user. This is the first step.

#!/usr/bin/perl -w

use strict;

my $user=`whoami`;
my @ps=`ps aux |grep $user`;
#print @ps;
foreach(@ps){
  $_ =~ s!.*pts/\([0-9][0-9]*\).*!$1!;
  print $_;
}
exit 0
[download]

I see. You want something like:

for (@ps){
  m!.*pts/([0-9]+)! and print $1;
}
[download]

Update:

It'd be better without the .*, actually. With it, it would fail if the user in question were running a command that included the string pts/999999 because it would find the last one rather than the first.

Also, you might want to think about stuff like sshd: username@pts/1234 showing up in the commands. It's probably better to get the TTY from the right column rather than use a regex.

Update 2:

A couple quick points about what you were using and why not to do it that way...

$_ =~ s!.*pts/\([0-9][0-9]*\).*!$1!

Backwhacking your parens means to match actual parens, not to capture what's between them. Also, * means 0 or more and + means 1 or more, so [0-9][0-9]* is just a longer way of saying [0-9]+. Finally, there's no point in substituting out everything that doesn't match your captured parts when you can just print the captured part instead.

Oh, and you don't have to be explicit about working on $_ ... being able to use it implicitly is kind of its point.

-sauoq
"My two cents aren't worth a dime.";