Regex to mask characters except first two and last two characters in Java

You can use

Find: (?<=..).(?=..)
Replace: *

Details:

• (?<=..) - a positive lookbehind that requires any two chars other than line break chars immediately to the leftht of the current location
• . - any char other than a line break char
• (?=..) - a positive lookahead that requires any two chars other than line break chars immediately to the right of the current location.

See the regex demo.

Here are solutions for some languages:

• java - text.replaceAll("(?<=..).(?=..)", "*")
• javascript - text.replace(/(?<=..).(?=..)/g, "*")
• python - re.sub(r"(?<=..).(?=..)", "*", text)
• c# - Regex.Replace(text, @"(?<=..).(?=..)", "*")
• vb.net - Regex.Replace(text, "(?<=..).(?=..)", "*")
• ruby - text.gsub(/(?<=..).(?=..)/, "*")
• r - gsub("(?<=..).(?=..)", "*", text, perl=TRUE)
• php - preg_replace('~(?<=..).(?=..)~', '*', $text)
• perl - $text =~ s/(?<=..).(?=..)/*/g;
• postgresql - REGEXP_REPLACE(text, '(?<=..).(?=..)', '*','g')

The following code use the regex (.{2})(.*)(.{2}) to do exactly what you asks (works with string from 5 chars above):

import java.util.regex.Matcher;
import java.util.regex.Pattern;

String input = "1234567890";

final Pattern hidePattern = Pattern.compile("(.{1,2})(.*)(.{1,2})");
Matcher m = hidePattern.matcher(input);
String output;

if(m.find()) {
  l = m.group(2).length();
  output = m.group(1) + new String(new char[l]).replace("\0", "*") + m.group(3);
}

System.out.println(output); // 12******90

If you need to allow less then 5 chars (till 3 below does not make sense) use:

(.{1,2})(.+)(.{1,2})

\A.{2}(.+).{2}\z will return you what you want. Will not work if less than 5 characters. This is language dependent. My regex is in ruby. In ruby:

\A => start of string \z => end of string

find your language equivalents.