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我是个问题,我的UNIX代码有unix代码问题?有
#!/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
好的选择选项的作品,但如果我在1或2作为一个选项来创建用户名或密码输入,它再次把我带回到了选择选项。无论我做什么按它将始终显示选择选项
有人可以帮助我使用bash在unix中运行此代码。
谢谢
也看看`select`命令(和`PS3`)。 – 2010-12-06 20:22:49