Masking credit card number using regex

Question

I am trying to mask the CC number, in a way that third character and last three characters are unmasked.

For eg.. 7108898787654351 to **0**********351

I have tried (?<=.{3}).(?=.*...). It unmasked last three characters. But it unmasks first three also.

Can you throw some pointers on how to unmask 3rd character alone?

Answer 1

You can use this regex with a lookahead and lookbehind:

str = str.replaceAll("(?<!^..).(?=.{3})", "*");
//=> **0**********351

RegEx Demo

RegEx Details:

• (?<!^..): Negative lookahead to assert that we don't have 2 characters after start behind us (to exclude 3rd character from matching)
• .: Match a character
• (?=.{3}): Positive lookahead to assert that we have at least 3 characters ahead

Answer 2

I would suggest that regex isn't the only way to do this.

char[] m = new char[16];  // Or whatever length.
Arrays.fill(m, '*');
m[2] = cc.charAt(2);
m[13] = cc.charAt(13);
m[14] = cc.charAt(14);
m[15] = cc.charAt(15);
String masked = new String(m);

It might be more verbose, but it's a heck of a lot more readable (and debuggable) than a regex.

Answer 3

Here is another regular expression:

(?!(?:\D*\d){14}$|(?:\D*\d){1,3}$)\d

See the online demo

It may seem a bit unwieldy but since a credit card should have 16 digits I opted to use negative lookaheads to look for an x amount of non-digits followed by a digit.

• (?! - Negative lookahead
  • (?: - Open 1st non capture group.
    • \D*\d - Match zero or more non-digits and a single digit.
    • ){14} - Close 1st non capture group and match it 14 times.
  • $ - End string ancor.
  • | - Alternation/OR.
  • (?: - Open 2nd non capture group.
    • \D*\d - Match zero or more non-digits and a single digit.
    • ){1,3} - Close 2nd non capture group and match it 1 to 3 times.
  • $ - End string ancor.
  • ) - Close negative lookahead.
• \d - Match a single digit.

This would now mask any digit other than the third and last three regardless of their position (due to delimiters) in the formatted CC-number.

Answer 4

Apart from where the dashes are after the first 3 digits, leave the 3rd digit unmatched and make sure that where are always 3 digits at the end of the string:

(?<!^\d{2})\d(?=[\d-]*\d-?\d-?\d$)

Explanation

• (?<! Negative lookbehind, assert what is on the left is not
  • ^\d{2} Match 2 digits from the start of the string
• ) Close lookbehind
• \d Match a digit
• (?= Positive lookahead, assert what is on the right is
  • [\d-]* 0+ occurrences of either - or a digit
  • \d-?\d-?\d Match 3 digits with optional hyphens
• $ End of string
• ) Close lookahead

Regex demo | Java demo

Example code

String regex = "(?<!^\\d{2})\\d(?=[\\d-]*\\d-?\\d-?\\d$)";
Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
String strings[] = { "7108898787654351", "7108-8987-8765-4351"};

for (String s : strings) {
    Matcher matcher = pattern.matcher(s);
    System.out.println(matcher.replaceAll("*"));
}

Output

**0**********351
**0*-****-****-*351

Answer 5

Don't think you should use a regex to do what you want. You could use StringBuilder to create the required string

String str = "7108-8987-8765-4351";
StringBuilder sb = new StringBuilder("*".repeat(str.length()));

for (int i = 0; i < str.length(); i++) {
  if (i == 2 || i >= str.length() - 3) {
    sb.replace(i, i + 1, String.valueOf(str.charAt(i)));
  }
}

System.out.print(sb.toString());   // output: **0*************351