Variable containing multiple args with quotes in Bash

Question

I generate a bash variable containing all my args and those args contain spaces. When I launch a command with those args - eg. ls $args - quotes are not correctly interprete

Answer 1

Use eval this will first evaluate any expansions and quoting and then execute the resultant string as if it had been typed into the shell.

args="'$f1' '$f2'"
eval ls $args

eval will then be executing ls 'file n1' 'file n2'

Had a very similar problem, trying to pass arguments in variables sourced from /etc/default/ to start_stop_daemon in init scripts.

Answer 2

Use set to set your variables as positional parameters; then quoting will be preserved if you refer to them via "$@" or "$1", "$2", etc. Make sure to use double quotes around your variable names.

set -- "$f1" "$f2"
touch "$@"
ls "$@"
rm "$@"

Answer 3

This is probably the worst answer, but you can change IFS. This is the "internal field separator" and is equal to space+tab+newline by default.

#!/bin/sh
IFS=,
MAR="-n,my file"
cat $MAR

The script above will run cat. The first argument will be -n (numbered lines) and the second argument will be my file.

Answer 4

You might consider using an array for the args, something like this:

args=( "$f1" "$f2" )
ls "${args[@]}"

(The problem you're hitting at the moment is that once interpolation has happened there's no difference between intra- and inter- filename spaces.)