有unix代码问题？

Question

我的unix代码有问题

#!/bin/bash
while : ; do   
echo "SELECT OPTION"   
echo "-------------"   
echo "1- Create username"   
echo "2- Create password"   
echo "3- Delete username"   
echo "4- Exit"   
read -p "enter option 1 2 3 or 4:" option   
case option in      
1) read -p "Enter username:"        
 adduser $REPLY && echo "Username successfully entered"   ;;      
2) passwd && "Password successfully entered" ;;      
3) read -p "Enter user to be deleted: "         
deluser $REPLY && echo "User deleted"  ;;      
4) exit  ;;      
*) continue  ;;   
esac
done

确定选择选项有效，但如果我输入1或2作为创建用户名或密码的选项，它会再次将我带回选择选项。无论我做什么，它都会显示选择选项

有人可以帮助我使用bash在unix中运行此代码。

谢谢

Answer 1

尝试$option - option被解释为字符串。