写一个脚本 通过字符菜单方式添加和删除用户

#!/bin/bash
#
function menu(){
cat << EOF
`echo -e "\033[30;47m########################################\033[0m"`
`echo -e "\033[32mMy Menu\033[0m"`
`echo -e "\033[30;47m########################################\033[0m"`
*       `echo -e "\033[35m1)add a user\033[0m"`
*       `echo -e "\033[35m2)delete a user\033[0m"`
*       `echo -e "\033[35m3)quit\033[0m"`
*       `echo -e "\033[35m4)return main menu\033[0m"`
EOF
}

#
clear
. menu.sh
menu
while true;do
read -p "please inoput a option: " option
case $option in
1)
read -p "add a user: " name
read -p "set passwd for the user: " pass
useradd $name &>/dev/null
if [ $? -eq 0 ];then
echo -e "\033[32;40m$name is created succesufully!!!\033[0m"
echo $pass | passwd --stdin $name &>/dev/null
else
echo -e "\033[32;40muser $name is created failly!!!\033[0m"
fi
;;
2)
read -p "delete a user: " name
userdel -r $name &>/dev/null
if [ $? -eq 0 ];then
echo -e "\033[32;40m$name is deleted successfully!!!\033[0m"
else
echo -e "\033[32;40m$name is deleted failly!!!\033[0m"
fi
;;
3)
echo -e "\033[32;40mquit successfully!!!\033[0m"
break
;;
4)
clear
menu
;;
esac
done