bash 如何修剪 UNIX who 命令的输出？
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How to trim the output of the UNIX who Command?
提问by Mike Needham
I am working with an idea using the unix whocommand. As we all know, there does not seem to be a direct switch that gives just the username and line (terminal) info without the date and screen info... eg: the output is mneedham tty7...2012-02-19 11:26 (:0)
我正在使用 unixwho命令处理一个想法。众所周知，似乎没有直接开关只提供用户名和线路（终端）信息而没有日期和屏幕信息......例如：输出是mneedham tty7...2012-02-19 11:26 (:0)
What I am trying to get is just the mneedham tty7part. The solution needs to work no matter how long the username and terminal information.
我想要得到的只是mneedham tty7一部分。无论用户名和终端信息有多长，该解决方案都需要工作。
I tried using tr -s ' '(one space) like who | tr -s ' 'and that gave me one space between everything. Not quite what I was seeking. Tried cut -d" " -f1gets the username only. So I am hopeful someone can help me find the right command to get both bits of information.
我尝试使用tr -s ' '（一个空格）like who | tr -s ' '，这给了我所有内容之间的一个空格。不完全是我想要的。尝试cut -d" " -f1仅获取用户名。所以我希望有人能帮助我找到正确的命令来获取这两种信息。
Thanks.
谢谢。
回答by Andrew Clark
Using cut:
使用切割：
who | cut -d " " -f1,2
Using awk:
使用 awk：
who | awk '{ print , }'
