统计出/etc/passwd文件中其默认shell为非/s bin/nologin的用户个数,并将用户都显示出来

[root@centos7 data]#grep -v "/sbin/nologin" /etc/passwd
root:x:0:0:root:/root:/bin/bash
sync:x:5:0:sync:/sbin:/bin/sync
shutdown:x:6:0:shutdown:/sbin:/sbin/shutdown
halt:x:7:0:halt:/sbin:/sbin/halt
lanfeng:x:1000:1000:lanfeng:/home/lanfeng:/bin/bash
mageia:x:1100:1100::/home/linux:/bin/bash
slackware:x:2002:2019::/home/slackware:/bin/tcsh
首先利用grep -v ,查找文本中/sbin/nologin所在行,并输出
[root@centos7 data]#grep -v "/sbin/nologin" /etc/passwd | cut -d: -f1
root
sync
shutdown
halt
lanfeng
mageia
slackware
用cut命令以:分号为分隔符,取第1列,即用户名
原文地址:https://www.cnblogs.com/qianyuezhan/p/13284274.html