How do I escape special characters for a substitution in a Perl one-liner?

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

As discussed at perldoc perlre:

...Today it is more common to use the quotemeta() function or the "\Q" metaquoting escape sequence to disable all metacharacters' special meanings like this:

/$unquoted\Q$quoted\E$unquoted/

Beware that if you put literal backslashes (those not inside interpolated variables) between "\Q" and "\E", double-quotish backslash interpolation may lead to confusing results. If you need to use literal backslashes within "\Q...\E", consult "Gory details of parsing quoted constructs" in perlop.

You can also use a ' as the delimiter in the s/// operation to make everything be parsed literally:

my $text = '@';
$text =~ s'@'1';
print $text;

In your example, you can do (note the single quotes):

perl -pe 's/\b\Q@f&\E\b/new_value/g if m/param5/ and not /^ *#/'

Answer 2

The other answers have covered the question, now here's your meta-problem: Leaning Toothpick Syndrome. Its when the delimiter and escapes start to blur together:

s/\/foo\/bar\\/\/bar\/baz/

The solution is to use a different delimiter. You can use just about anything, but balanced braces work best. Most editors can parse them and you generally don't have to worry about escaping.

s{/foo/bar\\}{/bar/baz}

Here's your regex with braced delimiters.

s{\@d\&}{new_value}

Much easier on the eyeholes.

Answer 3

If you really want to avoid typing the \s, put your search string into a variable and then use that in your regex instead. You don't need quotemeta or \Q ... \E in that case. For example:

my $s = '@d&';
s/$s/new_value/g;

If you must use this in a one-liner, bear in mind that you will have to escape the $s if you use "s to contain your perl code, or escape the 's if you use 's to contain your perl code.

Answer 4

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