in reply to Simple Encryption question

($_ contains a line of the file)     substr( $_, 0, 16) =~ tr/0-9/AB\-C-HJK/; That will make the substitution directly in $_ . A substitution like this is not sturdy encryption, just an obscuration.

Update: Changed tr string to fit the spec.

After Compline,
Zaxo

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Re: Re: Simple Encryption question
by rman (Initiate) on Dec 29, 2003 at 21:00 UTC
    Thank you perlmonks for the help. substr works fine in my case. yeah, this is by no means a robust encryption. One more question: How do i skip the first and last lines of the file from this substitution? The first and last lines are headers and footers of the file and needs to be preserved. thanks. I am revisiting perl after a long time and am rusty. So, appreciate all your help.
[reply]
      It depends on how your header and trailer data look like. If they do not start with numerics, then it's easy...
      while (<$filehandle>) {
    tr/blah/blah/ if /^\d{16}/; # do the translation
    print;                      # output result
}

      [download]
      If your header and trailer look similar to the rest of the data, and if your file is small, you could read everything in to memory first...
      my @data = <$filehandle>;
print $data[0];     # print first line
for (1..$#data-1) { # skip first and last line
  $data[$_] =~ tr/blah/blah;
  print $data[$_];
}
print $data[-1];    # print last line

      [download]
      Otherwise it will get a bit messy.

[reply]
[d/l]
[select]

      If your file is small, just read it into an array first: 

      open F, $file or die $!;
my @data = <F>;
for my $i ( 0..$#data ) {
    if ( $i == 0 or $i == $#data ) {
        #first or last line so do whatever
        print $data[$i];
    }
    else {
        # not first or last line
        $data[$i] =~ s#^(\d{16})#$_ = $1; tr/0-9/A-J/; $_#e
        print $data[$i];
    }
}

      [download]

      cheers

      tachyon

[reply]
[d/l]
        Brilliant! This works fine.

        I am curious to know how this statement works:

        $data[$i] =~ s#^(\d{16})#$_ = $1; tr/0-9/A-J/; $_#e

        I understand some pieces of it:
        -> the s is for substitution; ^d16 is for first 16digits;
        -> tr does the conversion. # delimits;

        Can you please let me know, how the whole piece fits in to do the substitution.

        thanks
[reply]
[d/l]

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