One-line command to find all real users in Linux

哈喽大家好,我是咸鱼。
接触过 Linux 的小伙伴们都知道在 Linux (或者说类 Unix)中,有三种类型的用户:
- 超级用户(UID 为 0):即 root 用户,拥有最高权限。
- 系统用户(UID 为 1~999):系统内建用户,用于运行系统服务和守护进程。
- 普通用户(UID 为 1000~60000):为了让使用者能够使用 Linux 系统资源而建立的,用户新建的账号一般就是普通账号。
那么现在问题来了,如何快速找出 Linux 中的真实用户(root 用户和普通用户)?
正式开始之前,我们先介绍一个工具——getent。
getent 命令
getent是"get entries"的缩写,主要用于检索系统数据库中的记录信息,如/etc/passwd、/etc/shadow、/etc/group、/etc/hosts等。它可以根据指定的数据库类型和键来查询特定条目。
(1) 查看本地的主机文件(/etc/hosts)中包含的指定域名信息。
[root@localhost ~]# getent hosts 127.0.0.1 127.0.0.1 localhost localhost.localdomain localhost4 localhost4.localdomain4
(2) 查看本地密码文件(/etc/passwd)中指定用户信息。
[root@localhost ~]# getent passwd user1 user1:x:1000:1000::/home/user1:/bin/bash
(3) 从 /etc/group 数据库中检索指定组信息。
[root@localhost ~]# getent group user1 user1:x:1000:
我们回到问题上来:如何快速找出 Linux 中的真实用户(root 用户和普通用户)?
方法一:
getent passwd | awk -F: '$3 >= 1000 && $3 < 65344 || $3 == 0 {print $1}'(1) getent passwd 用于检索 /etc/passwd 数据库中的所有用户信息
(2) awk 命令:
- -F::指定字段分隔符为冒号 (:)
- $3 >= 1000 && $3 < 65344:表示判断第三个字段(用户 ID)大于或等于 1000 且小于 65344
- $3 == 0:表示判断第三个字段(用户 ID)等于 0
- {print $1}:打印第一个字段(用户名)
总结一下:这条命令将检索 /etc/passwd 数据库中的所有用户信息,并打印出用户 ID 大于或等于 1000 且小于 65344,或者用户 ID 等于 0 的所有用户的用户名。
但是有的小伙伴看到这么一长串的命令就犯难了,”我文本三剑客掌握的不是很好,这条命令尤其是里面的判断看的我头都晕了,还有没有更简单的方法呢?”
方法二:
getent passwd 0 {1000..60000} | awk -F: '{print $1}'- getent passwd 0 {1000..60000} 命令用于检索系统中用户 ID 为 0 以及范围从 1000 到 60000 的所有用户信息
- awk -F: ‘{print $1}:表示对前面输出的内容进行处理——打印第一个字段
这里补充一下:getent passwd 0 {1000..60000} 命令使用花括号扩展 ({1000..60000}) 来生成一个 0 和从 1000 到 60000 的数字序列。
然后,getent passwd 命令使用这些数字作为参数,以检索系统中用户 ID 为 0 和从 1000 到 60000 的用户信息。
The above is the detailed content of One-line command to find all real users in Linux. For more information, please follow other related articles on the PHP Chinese website!
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