Bash工具从文件中获取第n行 [英] Bash tool to get nth line from a file

问题描述

有没有一种规范"的方式来做到这一点?我一直在使用 head -n |tail -1 可以解决问题，但我一直想知道是否有专门从文件中提取一行(或一系列行)的 Bash 工具.

Is there a "canonical" way of doing that? I've been using head -n | tail -1 which does the trick, but I've been wondering if there's a Bash tool that specifically extracts a line (or a range of lines) from a file.

我所说的规范"是指一个主要功能就是这样做的程序.

By "canonical" I mean a program whose main function is doing that.

推荐答案

head 和带有 tail 的管道对于大文件来说会很慢.我建议 sed 像这样:

head and pipe with tail will be slow for a huge file. I would suggest sed like this:

sed 'NUMq;d' file

其中NUM是要打印的行号；因此，例如，sed '10q;d' file 打印 file 的第 10 行.

Where NUM is the number of the line you want to print; so, for example, sed '10q;d' file to print the 10th line of file.

说明:

NUMq当行号为NUM时会立即退出.

NUMq will quit immediately when the line number is NUM.

d 将删除该行而不是打印它；这在最后一行被禁止，因为 q 导致退出时跳过脚本的其余部分.

d will delete the line instead of printing it; this is inhibited on the last line because the q causes the rest of the script to be skipped when quitting.

如果变量中有 NUM，则需要使用双引号而不是单引号:

If you have NUM in a variable, you will want to use double quotes instead of single:

sed "${NUM}q;d" file

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