getopts的无参数提供

问题描述

如何检查是否有没有必需的参数前提？我发现：在开关外壳选项应该用于此目的是足够的，但它永远不会进入这种情况下（codeblock伪）。不要紧，我是否把冒号案开头或其他地方。

how to check whether there was no required argument provided? I found that ":" option in switch case should be sufficient for this purpose, but it never enters that case (codeblock). It doesn't matter whether I put "colon-case" at the beginning or elsewhere.

我的code：

while getopts :a:b: OPTION;
do
     case "$OPTION" in
         a)
             var1=$OPTARG
             ;;
         b)
             var2=$OPTARG
             ;;
         ?)
             exitScript "`echo "Invalid option $OPTARG"`" "5"
             ;;
         :)
             exitScript "`echo "Option -$OPTARG requires an argument."`" "5"
             ;;
         *)
             exitScript "`echo "Option $OPTARG unrecognized."`" "5"
             ;;
     esac
done

THX在前进。

THX in advance.

推荐答案

您必须逃离？。接下来就可以（部分）的作品。

You must escape the ?. The next can (partially) works.

err() { 1>&2 echo "$0: error $@"; return 1; }
while getopts ":a:b:" opt;
do
        case $opt in
                a) aarg="$OPTARG" ;;
                b) barg="$OPTARG" ;;
                :) err "Option -$OPTARG requires an argument." || exit 1 ;;
                \?) err "Invalid option: -$OPTARG" || exit 1 ;;
        esac
done

shift $((OPTIND-1))
echo "arg for a :$aarg:"
echo "arg for b :$barg:"
echo "unused parameters:$@:"

部分，因为当会调用上面的脚本

Partially because when will call the above script as

$ bash script -a a_arg -b b_arg extra

将工作像您期望的，

will works as you expect,

arg for a :a_arg:
arg for b :b_arg:
unused parameters:extra:

但是，当你称呼其为

But when you will call it as

bash script -a -b b_arg

将打印

arg for a :-b:
arg for b ::
unused parameters:b_arg:

什么不是，你想要什么。

what is not, what you want.

和UUOE。 （Useles使用echo）。

And UUOE. (Useles use of echo).

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