grep -w只有空格作为分隔符 [英] grep -w with only space as delimiter

问题描述

grep -w uses punctuations and whitespaces as delimiters. 

如何将grep设置为仅使用空格作为分隔符？

How can I set grep to only use whitespaces as a delimiter for a word?

推荐答案

c>与 grepfoo相同。如果您还想匹配行尾或制表符，您可以开始执行如下操作： grep'\（^ \ | \）foo\（$ \ | \）'，但你最好用 perl -ne'print if / \sfoo\s /'

If you want to match just spaces: grep -w foo is the same as grep " foo ". If you also want to match line endings or tabs you can start doing things like: grep '\(^\| \)foo\($\| \)', but you're probably better off with perl -ne 'print if /\sfoo\s/'

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