Perl - Convert integer to text Char(1,2,3,4,5,6) - perl

Question

I am after some help trying to convert the following log I have to plain text.
This is a URL so there maybe %20 = 'space' and other but the main bit I am trying convert is the char(1,2,3,4,5,6) to text.
Below is an example of what I am trying to convert.
select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)
What I have tried so far is the following while trying to added into the char(in here) to convert with the chr($2)
perl -pe "s/(char())/chr($2)/ge"
All this has manage to do is remove the char but now I am trying to convert the number to text and remove the commas and brackets.
I maybe way off with how I am doing as I am fairly new to to perl.
perl -pe "s/word to remove/word to change it to/ge"
"s/(char(what goes in here))/chr($2)/ge"
Output try to achieve is
select -x1-Q-,-x2-Q-,-x3-Q-
Or
select%20-x1-Q-,-x2-Q-,-x3-Q-
Thanks for any help

Answer 1

There's too much to do here for a reasonable one-liner. Also, a script is easier to adjust later
use warnings;
use strict;
use feature 'say';
use URI::Escape 'uri_unescape';
my $string = q{select%20}
. q{char(45,120,49,45,81,45),char(45,120,50,45,81,45),}
. q{char(45,120,51,45,81,45)};
my $new_string = uri_unescape($string); # convert %20 and such
my #parts = $new_string =~ /(.*?)(char.*)/;
$parts[1] = join ',', map { chr( (/([0-9]+)/)[0] ) } split /,/, $parts[1];
$new_string = join '', #parts;
say $new_string;
this prints
select -x1-Q-,-x2-Q-,-x3-Q-
Comments
Module URI::Escape is used to convert percent-encoded characters, per RFC 3986
It is unspecified whether anything can follow the part with char(...)s, and what that might be. If there can be more after last char(...) adjust the splitting into #parts, or clarify
In the part with char(...)s only the numbers are needed, what regex in map uses
If you are going to use regex you should read up on it. See
perlretut, a tutorial
perlrequick, a quick-start introduction
perlre, the full account of syntax
perlreref, a quick reference (its See Also section is useful on its own)

Answer 2

Alright, this is going to be a messy "one-liner". Assuming your text is in a variable called $text.
$text =~ s{char\( ( (?: (?:\d+,)* \d+ )? ) \)}{
my #arr = split /,/, $1;
my $temp = join('', map { chr($_) } #arr);
$temp =~ s/^|$/"/g;
$temp
}xeg;
The regular expression matches char(, followed by a comma-separated list of sequences of digits, followed by ). We capture the digits in capture group $1. In the substitution, we split $1 on the comma (since chr only works on one character, not a whole list of them). Then we map chr over each number and concatenate the result into a string. The next line simply puts quotation marks at the start and end of the string (presumably you want the output quoted) and then returns the new string.
Input:
select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)
Output:
select%20"-x1-Q-","-x2-Q-","-x3-Q-"
If you want to replace the % escape sequences as well, I suggest doing that in a separate line. Trying to integrate both substitutions into one statement is going to get very hairy.

Answer 3

This will do as you ask. It performs the decoding in two stages: first the URI-encoding is decoded using chr hex $1, and then each char() function is translated to the string corresponding to the character equivalents of its decimal parameters
use strict;
use warnings 'all';
use feature 'say';
my $s = 'select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)';
$s =~ s/%(\d+)/ chr hex $1 /eg;
$s =~ s{ char \s* \( ( [^()]+ ) \) }{ join '', map chr, $1 =~ /\d+/g }xge;
say $s;
output
select -x1-Q-,-x2-Q-,-x3-Q-