Bash:命令输出中的grep模式 [英] Bash: grep pattern from command output

问题描述

我对bash真的很陌生，但这是学校的主题之一.练习之一是:

I'm really new with bash, but it's one of the subjects on school. One of the exercises was:

提供文件"/etc/passwd"的行号，有关您自己的登录信息.

假设USERNAME是我自己的登录ID，我可以用这种方式完美地做到这一点:

Suppose USERNAME is my own login ID, I was able to do it perfectly in this way:

 cat /etc/passwd -n | grep USERNAME | cut -f1

只需给出所需的行号(可能有一种更优化的方法).但是，我想知道是否有一种方法可以使该命令更通用，以便它使用whoami的输出来表示grep模式，而不使用脚本或使用变量.换句话说，要使其成为易于阅读的单行命令，如下所示:

Which simply gave the line number required (there may be a more optimised way). I wondered however, if there was a way to make the command more general so that it uses the output of whoami to represent the grep pattern, without scripting or using a variable. In other words, to keep it an easy-to-read one-line command, like so:

 cat /etc/passwd -n | grep (whoami) | cut -f1

对不起，如果这是一个真正的菜鸟问题.

Sorry if this is a really noob question.

推荐答案

cat /etc/passwd -n | grep `whoami` | cut -f1 

将命令括在`标记中使它执行该命令，并将输出发送到包装的命令中.

Surrounding a command in ` marks makes it execute the command and send the output into the command it's wrapped in.

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