Setting variable at perl invocation

Question

I'm trying to set variable for the perl via the -s and -var=value.

perl -s -E 'say $xx if $xx' -xx=abc

prints

No Perl script found in input

It probably didn't works together with the -E or i'm doing something wrong. (the perlrun doesn't mentions about the wrong combination of -E and -s)

ysth · Accepted Answer · 2014-09-08 21:14:26Z

5

You need a -- to end the built-in perl options and start your user supplied ones:

perl -s -E 'say $xx if $xx' -- -xx=abc

As is, perl thinks you are using its -x switch.

(That said, -s is a really lousy thing to try to use these days; use Getopt::Long.)

answered Sep 8, 2014 at 21:14

ysth

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12 Comments

kobame Over a year ago

thanx - and yes, the perlrun mentions the -- but didnt understand how it means it. ;)

ysth Over a year ago

it doesn't do anything besides separate between perl switches and program name/arguments.

kobame Over a year ago

i like the -s idea for the one-liners.

user3885927 Over a year ago

I get this error that goes away when I use double quotes (I replaced say with print since I don't have a say function...): Can't find string terminator "'" anywhere before EOF at -e line 1.

ysth Over a year ago

@kobame: how about instead: XX=ABC perl -E 'say $ENV{XX} if $ENV{XX}'

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