perl regex 中的单独反向引用后跟数字文字 [英] Seperate backreference followed by numeric literal in perl regex

问题描述

我发现了这个相关的问题:在 perl 中，替换文本中的反向引用，后跟数字文字但似乎完全不同.我有一个这样的正则表达式

I found this related question : In perl, backreference in replacement text followed by numerical literal but it seems entirely different. I have a regex like this one

s/([^0-9])([xy])/\1 1\2/g
                   ^
              whitespace here

但是这个空格出现在替换中.

But that whitespace comes up in the substitution.

如何在不让 perl 混淆对 \11 的反向引用的情况下获取替换字符串中的空格?

How do I not get the whitespace in the substituted string without having perl confuse the backreference to \11?

例如.15+x+y 变为 15+1x+1y.我想得到 15+1x+1y.

For eg. 15+x+y changes to 15+ 1x+ 1y. I want to get 15+1x+1y.

推荐答案

\1 是一个正则表达式原子，它匹配第一个捕获捕获的内容.在替换表达式中使用它是没有意义的.你想要 $1.

\1 is a regex atom that matches what the first capture captured. It makes no sense to use it in a replacement expression. You want $1.

$ perl -we'$_="abc"; s/(a)/\1/'
\1 better written as $1 at -e line 1.

在字符串文字(包括替换的替换表达式)中，您可以使用卷曲分隔 $var:${var}.这意味着您需要以下内容:

In a string literal (including the replacement expression of a substitution), you can delimit $var using curlies: ${var}. That means you want the following:

s/([^0-9])([xy])/${1}1$2/g

以下更有效(尽管对 xxx 给出了不同的答案):

The following is more efficient (although gives a different answer for xxx):

s/[^0-9]\K(?=[xy])/1/g

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