Question

Is there some way to replace a string such as @or * or ? or & without needing to put a “\” before it?

Example:

perl -pe 'next if /^#/; s/\@d\&/new_value/ if /param5/' test

In this example I need to replace a @d& with new_value but the old value might contain any character, how do I escape only the characters that need to be escaped?

Answer 1

You have several problems:

1. You are using \b incorrectly
2. You are replacing code with shell variables
3. You need to quote metacharacters

From perldoc perlre

A word boundary (“\b”) is a spot between two characters that has a “\w” on one side of it

Neither of the characters @ or & are \w characters. So your match is guaranteed to fail. You may want to use something like s/(^|\s)\@d\&(\s|$)/${1}new text$2/

(^|\s) says to match either the start of the string (^)or a whitespace character (\s).

(\s|$) says to match either the end of the string ($) or a whitespace character (\s).

To solve the second problem, you should use %ENV.

To solve the third problem, you should use the \Q and \E escape sequences to escape the value in $ENV{a}.

Putting it all together we get:

#!/bin/bash

export a='@d&'
export b='new text'

echo 'param5 @d&' | 
    perl -pe 'next if /^#/; s/(^|\s)\Q$ENV{a}\E(\s|$)/$1$ENV{b}$2/ if /param5/' 

Which prints

param5 new text