linux - 从shell脚本中的字符串中删除单引号

Question

这是我的 shell 脚本-

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

当我运行脚本时，我得到以下输出-

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

问题是我不想用单引号('hi'、'hello'、'bye')显示参数。我应该怎么做才能去掉这些引号？

Answer 1

为getopt使用选项-u或--unquoted，即

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

getopt 的联机帮助页对 -u 说:

Do not quote the output.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).