CODEWITHSUNDEEP
linuxshell
AlwaysALearner
AlwaysALearner

Reputation: 6690

Remove single quotes from string in shell script

This is my shell script-

if ! options=$(getopt -o : -l along:,blong:,clong: -- "$@")
then
    # something went wrong, getopt will put out an error message for us
    exit 1
fi

set -- $options

while [ $# -gt 0 ]
do
    case $1 in
        --along) echo "--along selected :: $2" ;;
        --blong) echo "--blong selected :: $2" ;;
        --clong) echo "--clong selected :: $2" ;;
    esac
    shift
done

when i run the script i get the following output-

./test.sh --along hi --blong hello --clong bye
--along selected :: 'hi'
--blong selected :: 'hello'
--clong selected :: 'bye'

The problem is I don't want to display the arguments with single quotes ('hi', 'hello', 'bye'). What should I do to get rid of those quotes?

Upvotes: 1

Views: 2892

Answers (1)

Matthias
Matthias

Reputation: 7521

Use the option -u or --unquoted for getopt, i.e.

if ! options=$(getopt -u -o : -l along:,blong:,clong: -- "$@")

The manpage of getopt says for -u:

Do not quote the output. Note that whitespace and special (shell-dependent) characters can cause havoc in this mode (like they do with other getopt(1) implementations).

Upvotes: 6

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