如何防止将参数传递给在 getopt 中不带参数的参数？

Question

例如考虑这个脚本，

#!/bin/sh

OPTS=`getopt -o ahb:c: --long help,name:,email: -- "$@"`
#echo "$OPTS"
eval set -- "$OPTS"

usage () {
    echo "type -h for help"
}


while true; do
    case "" in
        -a) echo "a is a short option that do not take parameters"; shift ;;
        -b) echo "b is a short option that requires one parameter and you specified "; shift 2;;
        -c) echo "c is a short option that requires one parameter and you specified "; shift 2;;
        --name) echo "your name is "; shift 2;;
        --email) echo "your email is "; shift 2;;
        -h | --help) echo "Google search yourself !"; shift 1;;
        --) usage ; shift; break ;;
        *) echo "hello"; break ;;

    esac
done

因此，如果将脚本调用为 sh myscript.sh -a hello，它应该抛出一个错误，告知 -a 不接受任何参数。

有办法吗？

Answer 1

您遇到的问题是因为您不需要 "second" 参数但它实际上是第三个参数！

如果您不需要-a 后的任何参数，那么您需要检查$3 是否存在。 ($2 将是 '--')

这是固定代码，在 -a 之后提供内容时会打印错误：

#!/bin/sh

OPTS=`getopt -o ahb:c: --long help,name:,email: -- "$@"`
#echo "$OPTS"
eval set -- "$OPTS"

usage () {
    echo "type -h for help"
}


while true; do
    case "" in
        -a) if [ ! -z "" ] ; then 
#Check if there is something after -a
                echo "a is a short option that do not take parameters";
else
         echo "Whatever..";
fi
 shift ;;
        -b) echo "b is a short option that requires one parameter and you specified "; shift 2;;
        -c) echo "c is a short option that requires one parameter and you specified "; shift 2;;
        --name) echo "your name is "; shift 2;;
        --email) echo "your email is "; shift 2;;
        -h | --help) echo "Google search yourself !"; shift 2;;
        --) usage ; shift; break ;;
        *) echo "hello"; break ;;

    esac
done

希望这就是您要找的:)

如何防止将参数传递给在 getopt 中不带参数的参数？

How to prevent passing parameters to arguments that don't take one in getopt?

linux

bash

parsing

parameters

getopt