cut或awk命令以打印第一行的第一字段 [英] cut or awk command to print first field of first row

问题描述

我正在尝试打印输出第一行的第一字段.就是这种情况.我只需要从此输出中打印SUSE.

I am trying print the first field of the first row of an output. Here is the case. I just need to print only SUSE from this output.

# cat /etc/*release

SUSE Linux Enterprise Server 11 (x86_64)
VERSION = 11
PATCHLEVEL = 2

尝试过cat /etc/*release | awk {'print $1}'，但会打印每行的第一个字符串

Tried with cat /etc/*release | awk {'print $1}' but that print the first string of every row

SUSE
VERSION
PATCHLEVEL

推荐答案

指定

Specify NR if you want to capture output from selected rows:

awk 'NR==1{print $1}' /etc/*release

实现此目标的另一种方法( ugly )是:

An alternative (ugly) way of achieving the same would be:

awk '{print $1; exit}'

从特定行(例如第42行)获取第一个字符串的有效方法是:

An efficient way of getting the first string from a specific line, say line 42, in the output would be:

awk 'NR==42{print $1; exit}'

这篇关于cut或awk命令以打印第一行的第一字段的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！