sk found two problems, here's a couple more. Your mytool.pl line will not run without a path indication unless its directory is in your $PATH. Double quoting is undesirable since it will make the shell try to interpolate. This works for me,

$ cat >mytool.pl
#!/usr/bin/perl
print 'Substitute string';
$ chmod 755 mytool.pl
$ cat >test.tmpl
This is mytool.pl foo
$ perl -pe's/(mytool\.pl\b.*)/`.\/$1`/e' test.tmpl
This is Substitute string

$
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(Updated to thank the proper folks. :) Thanks for all the friendly reminders)

Excellent zaxo and sk! You make it look too easy. :)

zaxo, you are right about the double quotes and that is why I had the "\" before the "$1". In my original try I had single quotes, but there I saw that it wasn't executing, but rather just replacing with exactly what I wanted to execute. So, I want to run this from the cmd line and I'll mix up what the two of you both shared.

I finally went with this (mytool.pl is in my PATH):

perl -i.bak -pe 's/^(mytool\.pl .*)/`$1`/e' test.template

Nice. Thank you both!

Shannon Kerr

If you run it without /e you will notice the string becomes `printhi`. Now you want to eval this expression and the /e takes care of that.

This works for me

 
#!/usr/bin/perl

use strict;
use warnings;

my $str = "printmeprinthi";

$str =~ s/printme(.*)$/`$1`/e;

print $str,$/;
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my printhi program is

#!/bin/sh
echo hi
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output

hi
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