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javaregexmasking
Cassie
Cassie

Reputation: 3099

Masking phone number Java

I need to mask the phone number. it may consist of the digits, + (for country code) and dashes. The country code may consist of 1 or more digits. I have created such kind of regular expression to mask all the digits except the last 4:

inputPhoneNum.replaceAll("\\d(?=\\d{4})", "*");

For such input: +13334445678

I get this result: +*******5678

However, it doesn't work for such input: +1-333-444-5678 In particular, it returns just the same number without any change. While the desired output is masking all the digits except for the last 4, plus sign and dashes. That is why I was wondering how I can change my regular expression to include dashes? I would be grateful for any help!

Upvotes: 11

Views: 26392

Answers (6)

Winter MC
Winter MC

Reputation: 375

This is what I used, it may be useful, just masks some digits in the provided number

 /*
 * mask mobile number .
 */
public  String maskMobileNumber(String mobile) {
    final String mask = "*******";
    mobile = mobile == null ? mask : mobile;
    final int lengthOfMobileNumber = mobile.length();
    if (lengthOfMobileNumber > 2) {
        final int maskLen = Math.min(Math.max(lengthOfMobileNumber / 2, 2), 6);
        final int start = (lengthOfMobileNumber - maskLen) / 2;
        return mobile.substring(0, start) + mask.substring(0, maskLen) + mobile.substring(start + maskLen);
    }
    return mobile;
}

Upvotes: 0

achAmháin
achAmháin

Reputation: 4266

If you don't want to use regex, an alternate solution would be to loop through the String with a StringBuilder from end to start, and append the first 4 digits and then * after that (and just append any non-digit characters as normal)

public static String lastFour(String s) {
        StringBuilder lastFour = new StringBuilder();
        int check = 0;
        for (int i = s.length() - 1; i >= 0; i--) {
            if (Character.isDigit(s.charAt(i))) {
                check++;
            }
            if (check <= 4) {
                lastFour.append(s.charAt(i));
            } else {
                lastFour.append(Character.isDigit(s.charAt(i)) ? "*" : s.charAt(i));
            }
        }
        return lastFour.reverse().toString();
    }

Try it online!

Upvotes: 0

anubhava
anubhava

Reputation: 785276

Use this regex for searching:

.(?=.{4})

RegEx Demo

Difference is that . will match any character not just a digit as in your regex.

Java code:

inputPhoneNum = inputPhoneNum.replaceAll(".(?=.{4})", "*");

However if your intent is to mask all digits before last 4 digits then use:

.(?=(?:\D*\d){4})

Or in Java:

inputPhoneNum = inputPhoneNum.replaceAll("\\d(?=(?:\\D*\\d){4})", "*");

(?=(?:\\D*\\d){4}) is a positive lookahead that asserts presence of at least 4 digits ahead that may be separated by 0 or more non-digits.

RegEx Demo 2

Upvotes: 18

Harshvinder Singh
Harshvinder Singh

Reputation: 21

I think this should work

".*\\d(?=\\d{4})","*"

You can try creating by hit and trial using this website.

Upvotes: 0

Youcef LAIDANI
Youcef LAIDANI

Reputation: 59985

Try to use two replace all non digit or + with empty then use your regex :

"+1-333-444-5678".replaceAll("[^\\d\\+]", "").replaceAll("\\d(?=\\d{4})", "*");

Output

+*******5678

Upvotes: 2

Mustapha Belmokhtar
Mustapha Belmokhtar

Reputation: 1219

I'm not good in RegEx but I think you should normalize the phone numbers by getting rid of -occurences : 

   inputPhoneNum = inputPhoneNum.replace("-","").replaceAll("\\d(?=\\d{4})", "*");

Upvotes: 2

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