如何将用户输入读入 Bash 中的变量?

本文介绍了如何将用户输入读入 Bash 中的变量?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在尝试创建一个脚本来简化在 iOS 设备上创建新用户的过程.以下是分解步骤.

I'm trying to create a script that simplifies the process of creating a new user on an iOS device. Here are the steps broken down.

fullname="USER INPUT" user="USER INPUT" group=$user uid=1000 gid=1000 home=/var/$user echo "$group:*:$gid:$user" >> /private/etc/group echo "$user::$uid:$gid::0:0:$fullname:$home:/bin/sh" >> /private/etc/master.passwd passwd $user mkdir $home chown $user:$group $home

如您所见，某些字段需要输入.如何在脚本中请求输入变量?

As you can see some fields require input. How can I request input for a variable in script?

推荐答案

使用read -p:

# fullname="USER INPUT" read -p "Enter fullname: " fullname # user="USER INPUT" read -p "Enter user: " user

如果您想确认:

read -p "Continue? (Y/N): " confirm && [[ $confirm == [yY] || $confirm == [yY][eE][sS] ]] || exit 1

您还应该引用您的变量以防止路径名扩展和用空格进行分词:

You should also quote your variables to prevent pathname expansion and word splitting with spaces:

# passwd "$user" # mkdir "$home" # chown "$user:$group" "$home"