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using sed buffers

by floobit (Initiate)
on May 02, 2012 at 21:31 UTC ( [id://968550]=perlquestion: print w/replies, xml ) Need Help??

floobit has asked for the wisdom of the Perl Monks concerning the following question:

This question has probably been posted before, but I have spent about 2 hours looking through the doc and searching this forum and found nothing. I apologize:

In linux, I am trying to strip the tty off a ps aux. Here is the regular expression I would use in sed. It works wonderfully:


When used in perl, it does not work. I have narrowed it down to the use of \( and \). When I remove these, the regex actually does something. How do I use the buffer functionality of sed in perl? Does anyone know a good online reference explaining the differences of perl's sed and canonical sed?

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Re: using sed buffers
by ww (Archbishop) on May 02, 2012 at 21:37 UTC
    Don't use sed; use a regex, re which, see perlrequick, perlretut and the regex section of our Tutorials.

    Also, please use code tags (<c>...</c) around code, as directed around the text-entry box you used.

    Update: Reading Markup in the Monastery will also help resolve your display issues.

    However, a hint: perl regexen use $1, $2, etc for the buffers you seem to be seeking.

      I think his intention is to write it in Perl. After all, he's asking "this is what I do in sed, how do I translate it to perl".
      Thanks, I'll look those up. Is sed generally discouraged in perl?

        Actually, the point is that sed isn't a Perl function.

        What's discouraged (much of the time) is shelling out to a (legacy) OS-function because doing so is often more costly than using the Perl equivalent/variant.
Re: using sed buffers
by sauoq (Abbot) on May 02, 2012 at 21:41 UTC

    So, your code probably actually looks like this:

    Right? One point, s/// isn't "sed" ... it's "substitute". Now what do you really hope to do? And how are you calling it? If this you are doing this on a command line, how are you calling it exactly? ('Cuz there are better ways to quote than with all the slashes.) And what do you really mean that you want when you say "strip the tty off"?

    "My two cents aren't worth a dime.";
      One point, s/// isn't "sed" ... it's "substitute".
      Perl, of course, got its s/// from sed...
        Perl, of course, got its s/// from sed...

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

        "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

        I see. You want something like:

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


        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.

        "My two cents aren't worth a dime.";
Re: using sed buffers
by Anonymous Monk on May 03, 2012 at 05:12 UTC

    s2p helps you migrate from sed to perl

    $ s2p "s!.*pts/\([0-9][0-9]*\).*!\1!" produces

    ... BOS:; # s!.*pts/\([0-9][0-9]*\).*!\1! { $s = s !.*pts/(\d\d*).*!${1}!s; $CondReg ||= $s; }

     ${1} is $1 is \1
     \(\) is ()
    [0-9] is \d
    * is *
    . is .
      Brilliant. Thanks, all.

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