Prevent ssh from breaking up shell script parameters - linux

Question

I have a script, which is essentially a wrapper around an executable by the same name on a different machine. For the sake of example, i'll wrap printf here. My current script looks like this:
#!/bin/bash
ssh user#hostname.tld. printf "$#"
Unfortunately, this breaks when one of the arguments contains a space, e.g. i'd expect the following commands to give identical outputs.:
~$ ./wrap_printf "%s_%s" "hello world" "1"
hello_world1_
~$ printf "%s_%s" "hello world" "1"
hello world_1
The problem gets even worse when (escaped) newlines are involved. How would I properly escape my arguments here?

Answer 1

Based on the answer from Peter Lyons, but also allow quotes inside arguments:
#!/bin/bash
QUOTE_ARGS=''
for ARG in "$#"
do
ARG=$(printf "%q" "$ARG")
QUOTE_ARGS="${QUOTE_ARGS} $ARG"
done
ssh user#hostname.tld. "printf ${QUOTE_ARGS}"
This works for everything i've tested so far, except newlines:
$ /tmp/wrap_printf "[-%s-]" "hello'\$t\""
[-hello'$t"-]

Answer 2

#!/bin/sh
QUOTE_ARGS=''
for ARG in "$#"
do
QUOTE_ARGS="${QUOTE_ARGS} '${ARG}'"
done
ssh user#hostname.tld. "${QUOTE_ARGS}"
This works for spaces. It doesn't work if the argument has an embedded single quote.

Answer 3

Getting quoting right is pretty hard and doing it in bash (in a general and robust way) almost impossible.
Use Perl:
#!/usr/bin/perl
use Net::OpenSSH;
my $ssh = Net::OpenSSH->new('user#hostname');
$ssh->system('printf', #ARGV);

Answer 4

Based on the answers from Koert and Peter Lyons, here a wrapper for ssh; i call it "sshsystem". (also available at https://gist.github.com/4672115)
#!/bin/bash
# quote command in ssh call to prevent remote side from expanding any arguments
# uses bash printf %q for quoting - no idea how compatible this is with other shells.
# http://stackoverflow.com/questions/6592376/prevent-ssh-from-breaking-up-shell-script-parameters
sshargs=()
while (( $# > 0 )); do
case "$1" in
-[1246AaCfgKkMNnqsTtVvXxYy])
# simple argument
sshargs+=("$1")
shift
;;
-[bcDeFIiLlmOopRSWw])
# argument with parameter
sshargs+=("$1")
shift
if (( $# == 0 )); then
echo "missing second part of long argument" >&2
exit 99
fi
sshargs+=("$1")
shift
;;
-[bcDeFIiLlmOopRSWw]*)
# argument with parameter appended without space
sshargs+=("$1")
shift
;;
--)
# end of arguments
sshargs+=("$1")
shift
break
;;
-*)
echo "unrecognized argument: '$1'" >&2
exit 99
;;
*)
# end of arguments
break
;;
esac
done
# user#host
sshargs+=("$1")
shift
# command - quote
if (( $# > 0 )); then
# no need to make COMMAND an array - ssh will merge it anyway
COMMAND=
while (( $# > 0 )); do
arg=$(printf "%q" "$1")
COMMAND="${COMMAND} ${arg}"
shift
done
sshargs+=("${COMMAND}")
fi
exec ssh "${sshargs[#]}"

Answer 5

The easiest and quickest is to just use Bash's Quoting Parameter Transformation: ${parameter#Q}. This can automatically applied during array expansion with ${array[#]#Q}, but when using the builtin argument array, the name and the brackets are dropped, so it becomes ${##Q}. Therefore the original script only needs 4 characters added to it to work.
#!/bin/bash
ssh user#hostname.tld. printf "${##Q}"
Now any escaping will work, even terminal colors like this:
./wrap_printf "%s\e[39m\e[49m\n" $'\e[30m\e[42mBlack on Green' "Just Normal Text"