Re: remove chars with regex

[johngg]

An alternative to using a regex is substr applied at either end of the string.

johngg@aleatico:~$ perl -Mstrict -Mwarnings -E 'say q{};
my $char = q{0x388c818ca8b9251b393131c08a736a67ccb19297};
say join q{ ... },
   substr( $char, 0, 6 ),
   substr( $char, length( $char ) - 6, 6 );'

0x388c ... b19297

I hope this is helpful.

Update: Re. AnonyMonk's comment regarding negative offsets, I had completely forgotten that, how dumb am I :-(

Update 2: Here's the amended code.

johngg@aleatico:~$ perl -Mstrict -Mwarnings -E 'say q{};
my $char = q{0x388c818ca8b9251b393131c08a736a67ccb19297};
say join q{ ... },
   substr( $char, 0, 6 ),
   substr( $char, -6 );'

0x388c ... b19297

[Anonymous Monk]

FWIW, the second substr ... (for the last 6 characters) could be substr( $char, -6 ). Negative offsets are from the end, and if you omit the length you get all of the string after the start position.