Bash脚本参数

Question

我试图将参数传递给我编写的脚本，但无法正确理解。

我要的是一个没有标志一个强制的说法，并与国旗两个可选参数，因此它可以被称为是这样的：

./myscript mandatory_arg -b opt_arg -a opt_arg 

或

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

我看着getopts的和得到这个：

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

假设你强制性参数出现最后，那么你应该尝试下面的代码：[评论在线]

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 

---

如果你真的想要mandato RY参数出现第一，那么你可以做以下的事情：

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