jcc has asked for the wisdom of the Perl Monks concerning the following question:
I've tried other configurations:$ perl -se 'print $foo' -foo=bar Unrecognized switch: -foo=bar (-h will show valid options).
doesn't do anythingperl -e -s 'print $foo' -foo=bar
I know it's not a quoting issue:$ perl -es 'print $foo' -foo=bar Substitution pattern not terminated at -e line 1. $ perl -e -s -foo=bar 'print $foo' Unrecognized switch: -foo=bar (-h will show valid options).
Any thoughts?$ perl -e 'print $=' 60
|
|---|
| Replies are listed 'Best First'. | |
|---|---|
|
Re: -s switch with -e
by lidden (Curate) on Apr 17, 2006 at 20:43 UTC | |
|
Re: -s switch with -e
by ikegami (Patriarch) on Apr 17, 2006 at 22:27 UTC | |
|
Re: -s switch with -e
by Tanktalus (Canon) on Apr 17, 2006 at 20:44 UTC | |
|
Re: -s switch with -e
by tempest (Sexton) on Apr 17, 2006 at 20:48 UTC | |
by jcc (Sexton) on Apr 18, 2006 at 01:39 UTC |