如何在Linux bash中提取某个字符出现在第n个和第m个之间的字符串? [英] How to extract strings between nth and mth occurence of a certain character in linux bash?

问题描述

文件1包含:

a:b:c:d:any words here:e:f:G

w/r此处有任何单词"可以是一个单词，两个单词，三个单词，依此类推.

w/r "any words here" can be a single word, two words, three words, and so on.

我想获取第4个:"和第5个:"之间的字符串.因此，这将是这里的任何单词".

I want to get the string between the 4rd ":" and the 5th ":". So, that will be "any words here".

我最初的想法是用空格替换:"，然后使用awk打印..但是由于我要提取的字符串可以由多个单词组成，因此无法正常工作.

My initial idea was to replace ":" with space then, use awk to print.. but since the string i want to extract can be composed of multiple words, it will not accurately work.

推荐答案

cut 命令允许您基于定界符分割行，并从中提取必填字段

cut command allow you to split a line based on a delimiter, and extract required fields from it

在您的示例中，

> echo 'a:b:c:d:any words here:e:f:G' |cut -f 5 -d:

应该给你

any words here

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