Sed无法将非空白字符与字符类匹配 [英] Sed failed to match non whitespace characters with character class

问题描述

我想提取在/etc/lvm/lvm.conf 中配置的过滤规则，例如 filter = ["r |/dev/sda |"] .我希望 sed 返回"r |/dev/sda |" .因此，我尝试了以下脚本:

I want to extract filter rules configured in /etc/lvm/lvm.conf, like filter = [ "r|/dev/sda|" ]. I want sed to return "r|/dev/sda|". So I have tried the following script:

echo'filter = ["r |/dev/sda |"]'|sed -r's:^ \ s * filter \ s * = \ s * \ [\ s *([^ \ s] +)\ s * \]:\ 1:g'

但是它没有用，脚本返回了 filter = ["r |/dev/sda |"] .我已经尝试了一些在线正则表达式测试器，该组已正确匹配.

But it didn't work, the script has returned filter = [ "r|/dev/sda|" ]. I've tried a few on line regex tester, the group has been matched correctly.

但是，如果我将 [^ \ s] + 替换为.+ ，则可以.

However, if I replace [^\s]+ by .+, it works.

不是 [^ \ s] + 表示多个非空白字符吗?

请问有什么想法吗?

推荐答案

Acc.到 regular-expressions.info :

一个关键的语法差异是反斜杠不是POSIX括号表达式中的元字符.因此，在POSIX中，正则表达式 [\ d] 匹配 \ 或 d .

因此，您需要将 [^ \ s] 替换为 [^ [:space:]] (除空格以外的任何字符)

So you need to replace [^\s] with [^[:space:]] (any char other than whitespace).

示例:

echo ' filter = [ "r|/dev/sda|" ] ' | sed -E 's:^\s*filter\s*=\s*\[\s*([^[:space:]]+)\s*\]:\1:g'

输出:"r |/dev/sda |"

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