如何在Linux BASH中获得命令输出的一部分? [英] How do I get a part of the output of a command in Linux BASH?

问题描述

正如标题所述，如何在Bash中获得命令输出的一部分?

As the title says, how do I get a part of the output of a command in Bash?

例如，命令php -v输出:

PHP 5.3.28 (cli) (built: Jun 23 2014 16:25:09) 
Copyright (c) 1997-2013 The PHP Group
Zend Engine v2.3.0, Copyright (c) 1998-2013 Zend Technologies with the ionCube PHP Loader v4.6.1, Copyright (c) 2002-2014, by ionCube Ltd.

，而我只想输出PHP 5.3.28 (cli)部分，我该怎么做?

and I only want to output the PHP 5.3.28 (cli) part, how do I do that?

我尝试了php -v | grep 'PHP 5.3.28'，但是输出:PHP 5.3.28 (cli) (built: Jun 23 2014 16:25:09)，那不是我想要的.

I've tried php -v | grep 'PHP 5.3.28' but that outputs: PHP 5.3.28 (cli) (built: Jun 23 2014 16:25:09) and that's not what I want.

推荐答案

您可以尝试以下awk命令，

You could try the below awk command,

$ php -v | awk 'NR==1{print $1,$2,$3}'
PHP 5.3.28 (cli)

它从输入的第一行开始打印前三列.

It prints the first three columns from the first line of input.

• NR==1(条件)，即，仅当NR变量的值为1时，才执行{}中的语句.
• {print $1,$2,$3}打印col1，col2，col3.打印语句中的,表示OFS(输出字段分隔符)

• NR==1 (condition)ie, execute the statements within {} only if the value of NR variable is 1.
• {print $1,$2,$3} Print col1,col2,col3. , in the print statement means OFS(Output Field Seperator)

这篇关于如何在Linux BASH中获得命令输出的一部分?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！