庆典工具从文件中获得第n行 [英] bash tool to get nth line from a file

问题描述

是否有这样做的一个规范的方式？我一直在使用头-n |尾-1 这是卓有成效的，但如果有从一个文件一个bash工具，专门提取线（或行范围）我一直在想。

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.

由规范我的意思，其主要功能是这样做的。一个节目

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

推荐答案

头和管道与尾将慢一个巨大的文件。我建议 SED 是这样的：

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 是要打印行数;因此，例如， sed的'10Q; D'文件打印 10号线文件

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 当行号将立即退出 NUM 。

D 将删除行而不是打印出来;这是最后一行抑制，因为①导致脚本的其余部分退出的时候被跳过。

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.

如果您有 NUM 在一个变量，你将要使用双引号代替单：

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

sed "${NUM}q;d" file

这篇关于庆典工具从文件中获得第n行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！