BASH shell expand arguments with spaces from varia

Question

This question already has an answer here:

• Bash doesn't parse quotes when converting a string to arguments 4 answers

Say I have a variable $ARGS which contains the following:

file1.txt "second file.txt" file3.txt

How can I pass the contents of $ARGS as arguments to a command (say cat $ARGS, for example), treating "second file.txt" as one argument and not splitting it into "second and file.txt"?

Ideally, I'd like to be able to pass arguments to any command exactly as they are stored in a variable (read from a text file, but I don't think that's pertinent).

Thanks!

Answer 1

It's possible to do this without either bash arrays or eval: This is one of the few places where the behavior of xargs without either -0 or -d extensions (a behavior which mostly creates bugs) is actually useful.

# this will print each argument on a different line
# ...note that it breaks with arguments containing literal newlines!
xargs printf '%s\n' <<<"$ARGS"

...or...

# this will emit arguments in a NUL-delimited stream
xargs printf '%s\0' <<<"$ARGS"

# in bash 4.4, you can read this into an array like so:
readarray -t -d '' args < <(printf '%s\0' "$ARGS")
yourprog "${args[@]}" # actually run your programs

# in bash 3.x or newer, it's just a bit longer:
args=( );
while IFS= read -r -d '' arg; do
    args+=( "$args" )
done < <(xargs printf '%s\0' <<<"$ARGS")
yourprog "${args[@]}" # actually run your program

# in POSIX sh, you can't safely handle arguments with literal newlines
# ...but, barring that, can do it like this:
set --
while IFS= read -r arg; do
    set -- "$@" "$arg"
done < <(printf '%s\n' "$ARGS" | xargs printf '%s\n')
yourprog "$@" # actually run your program

...or, letting xargs itself do the invocation:

# this will call yourprog with ARGS given
# ...but -- beware! -- will cause bugs if there are more arguments than will fit on one
# ...command line invocation.
printf '%s\n' "$ARGS" | xargs yourprog

Answer 2

As mentioned by Jonathan Leffler you can do this with an array.

my_array=( "file1.txt" "second file.txt" "file3.txt" )
cat "${my_array[1]}"

An array's index starts at 0. So if you wanted to cat the first file in your array you would use the index number 0. "${my_array[0]}". If you wanted to run your command on all elements, replace the index number with @ or *. For instance instead of "${my_arryay[0]}" you would use "${my_array[@]}"Make sure you quote the array or it will treat any filename with spaces as separate files.

Alternatively if for some reason quoting the array is a problem, you can set IFS (which stands for Internal Field Separator) to equal a newline. If you do this, it's a good idea to save the default IFS to a variable before changing it so you can set it back to the way it was once the script completes. For instance:

# save IFS to a variable    
old_IFS=${IFS-$' \t\n'}
#set IFS to a newline
IFS='$\n'

# run your script
my_array=( "file1.txt" "second file.txt" "file3.txt" )
cat ${my_array[1]}

# restore IFS to its default state
IFS=$old_IFS

It's probably better to not mess around with IFS unless you have to. If you can quote the array to make your script work then you should do that.

For a much more in depth look into using arrays see:

• http://mywiki.wooledge.org/BashGuide/Arrays
• http://wiki.bash-hackers.org/syntax/arrays
• http://mywiki.wooledge.org/BashFAQ/005

Answer 3

Without bashisms, plain shell code might need an eval:

# make three temp files and list them.
cd /tmp ;  echo ho > ho ; echo ho ho > "ho ho" ; echo ha > ha ; 
A='ho "ho ho" ha' ; eval grep -n '.' $A

Output:

ho:1:ho
ho ho:1:ho ho
ha:1:ha

---

Note that eval is powerful, and if not used responsibly can lead to mischief...