用前导减号替换尾随减号 [英] Replace a trailing minus with leading minus

问题描述

我正在尝试替换带有尾随减号的值，并用前导减号替换它们.我有这个在 SED 中不太适用的正则表达式，虽然我不明白为什么.

I'm trying to replace values which have a trailing minus sign and replace them with a leading minus sign. I've got this regex which doesn't quite work in SED though I don't understand why.

echo '8.8-' | sed 's/(\d+(?:\.\d+)?)-/-\\1/'

SED 返回 8.8-

SED returns 8.8-

我在 Notepad++ RegEx Help 插件中尝试了匹配 8.8- 并将子匹配 \1 标识为 8.8 的表达式

I tried the expression in Notepad++ RegEx Help plugin which matches the 8.8- and identifies the sub-match \1 as 8.8

推荐答案

\d 是 Perlism.Sed 指的是较旧的、Perl 之前的 POSIX 标准.你可能想要的是

The \d is a Perlism. Sed refers to the older, pre-Perl, POSIX standard. What you probably want is

echo '8.8-' | sed -r 's/([[:digit:]]+(\.[[:digit:]]+)?)-/-\1/'

[[:digit:]] 是编写 Perl 的 \d 的 POSIX 方式.

The [[:digit:]] is the POSIX way to write Perl's \d.

当然，正如其他人指出的那样，您可以只写 [0-9].

Of course, you can just write [0-9], instead, as others have noted.

顺便说一句，您需要 -r 来使您的括号随心所欲.

Incidentally, you need the -r to make your parentheses work as you like.

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