从任何 Linux 命令输出中省略第一行
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Omitting the first line from any Linux command output
提问by AabinGunz
I have a requirement where i'd like to omit the 1st line from the output of ls -latr "some path"Since I need to remove total 136from the below output
我有一个要求，我想从输出中省略第一行，ls -latr "some path"因为我需要total 136从下面的输出中删除
So I wrote ls -latr /home/kjatin1/DT_901_linux//autoInclude/system | tail -qwhich excluded the 1st line, but when the folder is empty it does not omit it. Please tell me how to omit 1st line in any linux command output
所以我写了ls -latr /home/kjatin1/DT_901_linux//autoInclude/system | tail -q其中排除了第一行，但是当文件夹为空时它不会省略它。请告诉我如何在任何 linux 命令输出中省略第一行
采纳答案by Fredrik Pihl
Pipe it to awk:
管道它到awk：
awk '{if(NR>1)print}'
or sed
或者 sed
sed -n '1!p'
回答by wkl
This is a quick hacky way: ls -lart | grep -v ^total.
这是一种快速的hacky方式：ls -lart | grep -v ^total.
Basically, remove any lines that start with "total", which in lsoutput should only be the first line.
基本上，删除所有以“total”开头的行，它在ls输出中应该只是第一行。
A more general way (for anything):
更通用的方式（对于任何事情）：
ls -lart | sed "1 d"
ls -lart | sed "1 d"
sed "1 d"means only print everything but first line.
sed "1 d"意味着只打印除第一行之外的所有内容。
回答by Donal Fellows
The tailprogram can do this:
该tail程序可以这样做：
ls -lart | tail -n +2
The -n +2means “start passing through on the second line of output”.
的-n +2装置“开始穿过上输出的第二行”。
回答by Jeff Ferland
ls -lart | tail -n +2 #argument means starting with line 2
