retaining quotes in bash correctly

Question

I am trying to pass arguments from a bash script to an executable and one of them contains spaces. I have been searching how to solve this, but I cannot find the right way to do it. Minimal example with a script called first and a script called second.

first script:

#!/bin/bash
# first script
ARGS="$@"
./second $ARGS

second script:

#!/bin/bash
# second script
echo "got $# arguments"

Now if I run it like this, I get the following results:

$ ./first abc def
got 2 args
$ ./first "abc def"
got 2 args
$ ./first 'abc def'
got 2 args

How can I make it so, that the second script also only receives one argument?

Answer 1

You can't do it using an intermediate variable. If you quote it will always pass 1 argument, if you don't you will lose the quotes.

However, you can pass the arguments directly if you don't use the variable like this:

./second "$@"

 

$ ./first abc def
got 2 arguments
$ ./first "abc def"
got 1 arguments

Alternately, you can use an array to store the arguments like this:

#!/bin/bash
# first script
ARGS=("$@")
./second "${ARGS[@]}"

Answer 2

IFS is your friend .

#!/bin/bash
# first script
ARGS="$@"
IFS=$(echo -en "\n\b")
./second $ARGS

IFS stands for Internal Field Separator ...