Better Algorithm to Encode This?

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Is there a better programmatical way doing this encoding. (I don't need a better encryption method, just a better algorithm for getting the same results)

    $STRING =~ tr/a-mN-ZA-Mn-z0-9.&@=/H-Mn-z0-9.&@=a-mN-ZA-G/;

    $STRING =~ s/=/.3d/g;
    $STRING =~ s/&/.26/g;
    $STRING =~ s/@/.40/g;
    $STRING =~ s/\\/.5c/g;
    $STRING =~ s/\s/.20/g;
    $STRING =~ s/%/.25/g;
    $STRING =~ s/\//.2f/g;
    $STRING =~ s/#/.23/g;
    $STRING =~ s/"/.22/g;
    $STRING =~ s/'/.27/g;
    $STRING =~ s/\+/.2b/g;
    $STRING =~ s/\|/.7c/g;
    $STRING =~ s/\?/.3f/g;

    return $STRING;
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Re: Better Algorithm to Encode This? by Sifmole (Chaplain) on Jun 12, 2001 at 22:31 UTC
You might be interested in something like this... `# Gleefully ripped from the CGI.pm escape method. $STRING =~ s/([^a-zA-Z0-9_.-])/lc sprintf(".%02x", ord($1))/eg; # Originally #$toencode =~ s/([^a-zA-Z0-9_.-])/uc sprintf("%%%02x", ord($1))/eg;` [download]	[reply] [d/l]
Re: Re: Better Algorithm to Encode This? by extremely (Priest) on Jun 13, 2001 at 05:54 UTC
You'll need to encode the "." period for that to work correctly!!! (I couldn't help but golf it down a little, sorry.) `## No Period in the char class! $STRING =~ s/([^a-zA-Z0-9_-])/lc sprintf(".%02x", ord($1))/eg; ## my encode $S =~ s/([^\w-])/sprintf".%02x",ord$1/eg; ## my decode $S =~ s/\.(..)/chr hex $1/eg;` [download] -- $you = new YOU; honk() if $you->love(perl)	[reply] [d/l]
Re: Better Algorithm to Encode This? by tachyon (Chancellor) on Jun 12, 2001 at 23:02 UTC
You could use these two lines. The second replaces all your s/// statements and puts them into a one liner `$STRING =~ tr/a-mN-ZA-Mn-z0-9.&@=/H-Mn-z0-9.&@=a-mN-ZA-G/; $STRING =~ s/([=&@\\\s%\/#"'\+\\|\?])/ sprintf ".%x", ord $1 /ge;` [download] The second regex looks for the chars you want to change to hex digits of their ascii values with a leading "." (I won't ask why you want to do this although I am intrigued :-) We capture these chars in $1 via the ([chars go here]) construct. We then use ord to get the decimal ascii value and sprintf to change them to hex (and also add that "."). We need the /g to do all occurrences and the /e to evaluate the second part of the regex and use the return value as the substitution value for $1. Hope this helps Cheers tachyon	[reply] [d/l]