Bash – Passing Arguments from a File to a Bash Script

bashscriptingshell-script

I've got this situation:

./
./myscript.sh
./arguments.txt
./test.sh

Inside myscript.sh, I have to run the file test.sh, passing to it the arguments contained inside arguments.txt.

myscript.sh is:

arguments=$(cat arguments.txt)
source test.sh $arguments

This works well if if arguments.txt contains at most one argument:

firstargument

The substitution is:

++ source test.sh 'firstargument'

But the problem is with two or more arguments. It does this:

++ source test.sh 'firstargument secondargument'

Also, I don't know in advance the number of arguments inside arguments.txt. There can be zero or more.

Best Answer

Assuming each line of arguments.txt represents a separate argument, with bash 4 you can read arguments.txt into an array using mapfile (each line from the file goes in as an array element, in sequence) and then pass the array to the command

mapfile -t <arguments.txt
source test.sh "${MAPFILE[@]}"

The advantage is that splitting on spaces embedded inside lines is avoided

With lower versions of bash

IFS=$'\n' read -ra arr -d '' <arguments.txt
source test.sh "${arr[@]}"

Related Solutions

Shell – Optional arguments after or before the mandatory arguments

If you're fine with the optional arguments being at the end, you can just do this:

foo=$1
bar=$2
baz=${3:-default value}

That will store the first two arguments in $foo and $bar. If a third argument was provided, it will be stored in $baz; otherwise it will default to default value. You can also just check if the third variable is empty:

if [ -z "$3" ]; then
    # Code for the case that the user didn't pass a third argument
fi

If you want the defaults at the beginning, the easiest way is probably to check the number of arguments passed to the script, which is stored in $#. For example:

case $# in
  2)
    # Put commands here for when the user passes two arguments
    # Could just be setting variables like above, or anything else you need
    ;;
  3)
    # Put commands here for when the user passes three arguments
    ;;
  *)
    echo "Wrong number of arguments passed"
    exit 1;;
esac

Switching on $# will work fine for either case, if you want error checking. If your script uses getopt-style arguments like the cut example, you could use that; see this Stack Overflow question for more information

Bash – Shift bash arguments from the right

If the list of positional parameters is:

$ set -- 0wer 1wdfg 2erty 333 4ffff 5s5s5

Then this will print the arguments without the last:

$ echo "${@:1:$#-1}"
0wer 1wdfg 2erty 333 4ffff

Of course, the parameters could be set to that as well:

$ set -- "${@:1:$#-1}"
$ echo $@
0wer 1wdfg 2erty 333 4ffff

That works in bash version 2.0 or above.

For other simpler shells, you need a (somewhat tricky) loop to remove the last parameter:

unset b; for a; do set -- "$@" ${b+"$b"}; shift; b="$a"; done

Best Answer

Related Solutions

Shell – Optional arguments after or before the mandatory arguments

Bash – Shift bash arguments from the right

Related Question