Linux查看当前登录用户并踢出用户的命令

1、查看当前登录用户

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[wilsh@lcl ~]$ whatis w

w (1) - Show who is logged on and what they are doing

[wilsh@lcl ~]$ w

09:49:30 up 1 day, 17:19, 4 users, load average: 0.00, 0.00, 0.00

USER TTY FROM LOGIN@ IDLE JCPU PCPU WHAT

root tty3 - 09:25 23:25 0.10s 0.08s -bash

root pts/0 192.168.105.188 09:32 9:38 0.02s 0.02s -bash

root pts/1 192.168.105.188 09:36 9:32 0.03s 0.02s -bash

wilsh pts/2 192.168.105.188 09:41 0.00s 0.00s 0.00s w

2、踢出当前在线的用户

a)pkill

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[wilsh@lcl ~]$ whatis pkill

pkill [pgrep] (1) - look up or signal processes based on name and other attributes

[wilsh@lcl ~]$ pkill -KILL -u wilsh #-u用来指定用户名

[wilsh@lcl ~]$ pkill -9 -u wilsh # -KILL -9都是可以的。

[wilsh@lcl ~]$ pkill -KILL -t pts/2

[wilsh@lcl ~]$ pkill -KILL -t /dev/pts/2 # 这种写法是错误的。因为在man page中明确说明,如果要使用-t参数,那么就不能带/dev/前辍。<span style="color:#FF99FF;"><span style="background-color: rgb(255, 255, 153);">#在centos6.7 kernel 3.10.28中测试,发现写完整的设备名并不起作用(如/dev/pts/2),本文是在kernel 2.6.32中写的,昨天晚上刚刚重启编译完3.10.28的内核,测试了一下又不可以了,郁闷。</span></span>

b)skill

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[wilsh@lcl ~]$ whatis skill

skill

(1) - send a signal or report process status

[wilsh@lcl ~]$ skill -9 -t pts/2

[wilsh@lcl ~]$ w

09:57:37 up 1 day, 17:28, 4 users, load average: 0.00, 0.00, 0.00

USER TTY FROM LOGIN@ IDLE JCPU PCPU WHAT

root tty3 - 09:25 31:32 0.10s 0.08s -bash

root pts/0 192.168.105.188 09:32 17:45 0.02s 0.02s -bash

root pts/1 192.168.105.188 09:36 17:39 0.01s 0.01s -bash

wilsh pts/2 192.168.105.188 09:57 0.00s 0.00s 0.00s w

[wilsh@lcl ~]$ skill -9 -u wilsh

总结

以上所述是小编给大家介绍的Linux查看当前登录用户并踢出用户的命令,希望对大家有所帮助,如果大家有任何疑问欢迎给我留言,小编会及时回复大家的!

原文链接:https://www.cnblogs.com/caicairui/archive/2018/05/04/8989156.html

本文链接:https://my.lmcjl.com/post/5733.html

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