问题
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
But that whitespace comes up in the substitution.
How do I not get the whitespace in the substituted string without having perl confuse the backreference to
\11?
For eg.
15+x+y changes to 15+ 1x+ 1y.
I want to get 15+1x+1y.
回答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.
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$2/g
The following is more efficient (although gives a different answer for xxx):
s/[^0-9]\K(?=[xy])/1/g
回答2:
Just put braces around the number:
s/([^0-9])([xy])/${1}1${2}/g
来源：https://stackoverflow.com/questions/18721037/seperate-backreference-followed-by-numeric-literal-in-perl-regex