I would like to use a regular expression to mask all but the first three alphanumeric characters of each word in a string using a mask character (such as "x"), so "1 Buckingham Palace Road, London" would become "1 Bucxxxxxxx Palxxx Roax, Lonxxx".
Keeping the first three characters is easily done using
s/\b(\w{0,3})(.*)\b/$1/g
but I cannot seem to figure out how to insert length($2) times the masking character instead of $2.
Thanks!
C#:
new Regex(@"(?<!\b.{0,2}).").Replace("1 Buckingham Palace Road, London", "x");
Since you say it's language-agnostic, I trust this can be easily ported into your language of choice...
Or, you could just get the length of $2 and fill the x's the old fashioned way.
Positive lookbehind, any word character with three word characters before it gets changed to an X:
s/(?<=\w{3})\w/$1x/g;
example perl script:
my $string = "1 Buckingham Palace Road, London";
$string =~ s/(?<=\w{3})\w/$1x/g;
print qq($string\n);
use warnings;
use strict;
my $string = "1 Buckingham Palace Road, London";
$string =~ s(
\b(\w{0,3})(\w*)\b
){
$1 . ( 'x' x length $2 )
}gex;
print $string, "\n";
