Re: arguments to perl -e

I've been trying to come up to a solution for this using the -s flag to enable me to pass an additional custom flag with the variable as a parameter to the -e code. Why can't I use something like

perl -e -s -foo=bar 'print $foo'
[download]

UPDATE:
Turns out I can... thanks all:

#!/bin/ksh
var=bar
perl -se 'print $foo' -- -foo=${var}
[download]

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Re^2: arguments to perl -e by graff (Chancellor) on Apr 08, 2006 at 06:42 UTC
I had seen the "-s" option before, when reading perlrun, but had never tried using it till seeing your post. From what I can tell (on macosx 10.4.6, perl 5.8.6), it seems to work only if the script is stored in a file, and has a shebang line with this option flag -- and even then, it doesn't seem to work as advertised, unless the script is executable (runnable without passing it as an arg to "perl"): `$ perl -s -foo=bar -e 'print $foo,$/ if $foo' Unrecognized switch: -foo=bar (-h will show valid options). # (and similarly with other arrangements of these args) $ cat junk.pl #!/usr/bin/perl -s print "$foo\n" if $foo; $ perl junk.pl $ perl junk.pl -foo 1 $ perl junk.pl -foo=bar 1 # (I thought that should have printed "bar", but # there must be some logical reason why it didn't.) $ chmod +x junk.pl $ junk.pl $ junk.pl -foo 1 $ junk.pl -foo=bar bar # at last!` [download] So -s is just not suitable for use with a one-liner script on the command line. Oh well. At least there are other ways of doing this sort of thing (as indicated by earlier replies). UPDATE: Turns out the one-liner command-line script approach actually does work if you do it like this: `$ perl -se 'print "$foo\n" if $foo' -- -foo 1 $ perl -se 'print "$foo\n" if $foo' -- -foo=bar bar` [download] (thanks to blockhead for pointing this out to me.) There was something about the double-dash arg in the perlrun description of "-s", but it didn't seem to apply to this facet of perl's behavior.	[reply] [d/l] [select]
Re^2: arguments to perl -e by Anonymous Monk on Sep 16, 2013 at 19:20 UTC
huh, I almost thought it is impossible to make `-s` work when perl is called with `-i`, `-p` and `-e` but it can! The only way it could work is like: `perl -p -i -s -e 's/haha/$i/ge' -- -i=gaga "${FILES[@]}"` Thanks for all suggestions that made me find out this.	[reply] [d/l]