第三个嵌套if简化成这样，没啥漏洞吧，求检测

#!/bin/bash

read -t 30 -p "num2:" num2

read -t 30 -p "operator:" ope

if [ -n $num1 -a -n $num2 -a -n $ope ]

then

test1=$(echo $num1| sed 's/[0-9]//g')

test2=$(echo $num2| sed 's/[0-9]//g')

if [ -z "$test1" -a -z "$test2" ]

then

if [ $ope=='+' -o $ope=='-' -o $ope=='*' -o $ope=='/' ]

then

res=$((num1 $ope num2))

else

echo "no operator"

exit 10

fi

else

echo "no numbers"

exit 11

fi

else

echo "no values"

exit 12

fi