Passing arguments to a command in Bash script with spaces

Question

I'm trying to pass 2 arguments to a command and each argument contains spaces, I've tried escaping the spaces in the args, I've tried wrapping in single quotes, I've tried escaping \" but nothing will work.

Here's a simple example.

#!/bin/bash -xv

ARG="/tmp/a b/1.txt"
ARG2="/tmp/a b/2.txt"

ARG_BOTH="\"$ARG\" \"$ARG2\""
cat $ARG_BOTH

I'm getting the following when it runs:

ARG_BOTH="$ARG $ARG2"
+ ARG_BOTH='/tmp/a\ b/1.txt /tmp/a\ b/2.txt'
cat $ARG_BOTH
+ cat '/tmp/a\' b/1.txt '/tmp/a\' b/2.txt
cat: /tmp/a\: No such file or directory
cat: b/1.txt: No such file or directory
cat: /tmp/a\: No such file or directory
cat: b/2.txt: No such file or directory

SiegeX · Accepted Answer · 2011-01-20 23:56:04Z

13

See http://mywiki.wooledge.org/BashFAQ/050

TLDR

Put your args in an array and call your program as myutil "${arr[@]}"

#!/bin/bash -xv

file1="file with spaces 1"
file2="file with spaces 2"
echo "foo" > "$file1"
echo "bar" > "$file2"
arr=("$file1" "$file2")
cat "${arr[@]}"

Output

file1="file with spaces 1"
+ file1='file with spaces 1'
file2="file with spaces 2"
+ file2='file with spaces 2'
echo "foo" > "$file1"
+ echo foo
echo "bar" > "$file2"
+ echo bar
arr=("$file1" "$file2")
+ arr=("$file1" "$file2")
cat "${arr[@]}"
+ cat 'file with spaces 1' 'file with spaces 2'
foo
bar

edited Jan 20, 2011 at 23:56

answered Jan 20, 2011 at 23:50

SiegeX

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1 Comment

maep Over a year ago

Is there a solution for regular posix shells like dash?

DigitalRoss · Accepted Answer · 2011-01-21 00:21:10Z

6

This might be a good use-case for the generic "set" command, which sets the top-level shell parameters to a word list. That is, $1, $2, ... and so also $* and $@ get reset.

This gives you some of the advantages of arrays while also staying all-Posix-shell-compatible.

So:

set "arg with spaces" "another thing with spaces"
cat "$@"

answered Jan 21, 2011 at 0:21

DigitalRoss

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Comments

zwol · Accepted Answer · 2011-01-20 23:57:57Z

5

The most straightforward revision of your example shell script that will work correctly is:

#! /bin/sh

ARG="/tmp/a b/1.txt"
ARG2="/tmp/a b/2.txt"

cat "$ARG" "$ARG2"

However, if you need to wrap up a whole bunch of arguments in one shell variable, you're up a creek; there is no portable, reliable way to do it. (Arrays are Bash-specific; the only portable options are set and eval, both of which are asking for grief.) I would consider a need for this as an indication that it was time to rewrite in a more powerful scripting language, e.g. Perl or Python.

answered Jan 20, 2011 at 23:57

zwol

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2 Comments

Dennis Williamson Over a year ago

Would you care to say why set could be a source of grief?

zwol Over a year ago

There's only one "$@", so it can't be used for more than one thing at a time. And set -- $VARIABLE; cmd "$@" does word splitting exactly the same way cmd $VARIABLE does, so that's no good. And you have to make sure you aren't getting one of the many other things set does by accident.

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