当第一个选项是bash中的路径时，getopts

我对bash脚本中的getopts有问题。基本上，我的脚本必须通过以下方式调用：

./myScript/path/to/a/文件夹-a-b

我的代码顶部有以下内容：

while getopts ":ab" opt; do
 case $opt in
  a) 
   variable=a
   ;;
  b)    
   variable=b
   ;;
  \?)
   echo "invalid option -$OPTARG"
   exit 0
 esac
done

echo "$variable was chosen"

现在，只要我在没有/path/to/a/folder的情况下调用我的脚本，它就可以工作……我怎样才能让它工作呢

非常感谢

如果必须在参数之前放置路径，请使用a来弹出第一个位置参数，并将其余的留给getopts

# Call as ./myScript /path/to/a/folder -a -b

path_argument="$1"
shift   # Shifts away one argument by default

while getopts ":ab" opt; do
 case $opt in
  a) 
   variable=a
   ;;
  b)    
   variable=b
   ;;
  \?)
   echo "invalid option -$OPTARG"
   exit 0
 esac
done

echo "$variable was chosen, path argument was $path_argument"

如前所述，更标准的答案是将非选项参数放在选项之后。更喜欢这种样式，因为它使脚本与内置POSIX选项解析更加一致

# Call as ./myScript -a -b /path/to/a/folder 

while getopts ":ab" opt; do
 case $opt in
  a) 
   variable=a
   ;;
  b)    
   variable=b
   ;;
  \?)
   echo "invalid option -$OPTARG"
   exit 0
 esac
done
shift $((OPTIND - 1))  # shifts away every option argument,
                       # leaving your path as $1, and every
                       # positional argument as $@
path_argument="$1"
echo "$variable was chosen, path argument was $path_argument"

---

将路径放在参数之后。脚本如何处理

/path/to/a/文件夹
？