如何确保用户输入有效的开关盒选项？

Question

我试图弄清楚如何解决此代码...

echo -n "Enter you complaint : ";

read complaint;

if [[ -z $complaint ]] || [[ $complaint != 'FIRE' ]] || [[ $complaint != 'INTOXICATION' ]] || [[ $complaint != "INJURY" ]] || [[ $complaint != "BREAKAGE" ]] ;then

    echo "Not a good option";

    exit 1;

fi

case $complaint in

    FIRE)  = $complaint=$(echo "FIRE");;

    INTOXICATION) = $complaint=$(echo "INTOXICATION");;

    INJURY) = $complaint=$(echo "INJURY");;

    BREAKAGE) = $complaint=$(echo "BREAKAGE");;

esac

是否可能知道：为什么即使我键入有效的内容也总是输入？

Answer 1

中的条件if有误，应为：if [[ -z $complaint ]] || [[ ( $complaint != 'FIRE' && $complaint != 'INTOXICATION' && $complaint != "BRIS" && $complaint != "BOBO" ) ]] ;then也就是说，您想进入ifif$complaint是null（-z $complaint）或与任何有效选项都不匹配的情况，因此它必须同时与所有选项都不同，因此应&&改为||。或者，您也可以在case...esac块中使用默认条件：case $complaint in    FIRE)  = $complaint=$(echo "FIRE");;    INTOXICATION) = $complaint=$(echo "INTOXICATION");;    INJURY) = $complaint=$(echo "INJURY");;    BREAKAGE) = $complaint=$(echo "BREAKAGE");;    *) echo "Not a good option"; exit 1;;esac