Bash 脚本参数

Question

我试图将参数传递给我编写的脚本，但无法正确传递。

我想要的是一个没有标志的强制参数，和两个有标志的可选参数，所以它可以这样调用:

./myscript mandatory_arg -b opt_arg -a opt_arg

或

./myscript mandatory_arg -a opt_arg
./myscript mandatory_arg -b opt_arg

我查看了 getopts 并得到了这个:

while getopts b:a: option
do
    case "${option}"
    in
        b) MERGE_BRANCH=${OPTARG};;
        a) ACTION=${OPTARG};;
    esac
done

if "$1" = ""; then
    exit
fi

echo "$1"
echo "$MERGE_BRANCH"
echo "$ACTION"

但它根本不起作用。

Answer 1

假设您的强制参数出现在最后，那么您应该尝试以下代码:[内联评论]

OPTIND=1
while getopts "b:a:" option
do
    case "${option}"
    in
        b) MERGE_BRANCH=${OPTARG};;
        a) ACTION=${OPTARG};;
    esac
done

# reset positional arguments to include only those that have not
# been parsed by getopts

shift $((OPTIND-1))
[ "$1" = "--" ] && shift

# test: there is at least one more argument left

(( 1 <= ${#} )) || { echo "missing mandatory argument" 2>&1 ; exit 1; };

echo "$1"
echo "$MERGE_BRANCH"
echo "$ACTION"

结果:

~$ ./test.sh -b B -a A test
test
B
A
~$ ./tes.sh -b B -a A
missing mandatory argument

---

如果您真的希望强制参数出现第一个，那么您可以执行以下操作:

MANDATORY="${1}"
[[ "${MANDATORY}" =~ -.* ]] && { echo "missing or invalid mandatory argument" 2>&1; exit 1; };

shift # or, instead of using `shift`, you can set OPTIND=2 in the next line   
OPTIND=1
while getopts "b:a:" option
do
    case "${option}"
    in
        b) MERGE_BRANCH=${OPTARG};;
        a) ACTION=${OPTARG};;
    esac
done

# reset positional arguments to include only those that have not
# been parsed by getopts

shift $((OPTIND-1))
[ "$1" = "--" ] && shift

echo "$MANDATORY"
echo "$MERGE_BRANCH"
echo "$ACTION"

结果如下:

~$ ./test.sh test -b B -a A
test
B
A
~$ ./tes.sh -b B -a A
missing or invalid mandatory argument