Masking phone number Java

Question

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!

Is the expected output for the second number +*-***-***-5678? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Nov 21, 2017 at 9:18
Is +1-23-45-67 a case that should be considered and how should it be masked if so? — Sebastian Proske
– Sebastian Proske, Commented Nov 21, 2017 at 9:21
Check this solution will help you. stackoverflow.com/questions/34675260/… — Mitul Maheshwari
– Mitul Maheshwari, Commented Nov 21, 2017 at 9:22
I meant .replaceAll("\\d(?=(?:\\D*\\d){4})", "*"), but no idea if it is what you need, really. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Nov 21, 2017 at 9:33
@SebastianProske this case should not be considered. The main idea is that there are ***-**** such digits in the end and the beginning with country code may differ — Cassie
– Cassie, Commented Nov 21, 2017 at 9:37

anubhava · Accepted Answer · 2017-11-21 10:08:35Z

18

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

edited Nov 21, 2017 at 10:08

answered Nov 21, 2017 at 9:18

anubhava

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8 Comments

AxelH Over a year ago

So 123-465-789 will mask everything but -789 instead of "all the digits except the last 4"

AxelH Over a year ago

I don't know every phone format but it is probably valid somewhere ;) here we have a format like 0123-45-67-89.

Sebastian Proske Over a year ago

You are still masking the initial + and all - in the number (though OP did not specify the correct behaviour for this).

anubhava Over a year ago

Yes because OP wrote: While the desired output is masking all the digits except for the last 4, plus sign and dashes

AxelH Over a year ago

I like that solution even if you just said that +, - and last 4 digit should still be visible ;) but I still like the solution. (and just need to replace . with \d to match that requirements)

|

Mustapha Belmokhtar · Accepted Answer · 2017-11-21 09:26:16Z

2

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})", "*");

edited Nov 21, 2017 at 9:26

answered Nov 21, 2017 at 9:23

Mustapha Belmokhtar

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2 Comments

Wiktor Stribiżew Over a year ago

You may remove hyphens without a regex, .replaceAll("\\-","") => .replace("-",""). BTW, there is no need escaping - in a regex when outside a character class.

AxelH Over a year ago

@mustabelMo not really, not using a Regex is simpler/faster.

Youcef LAIDANI · Accepted Answer · 2017-11-21 09:32:33Z

2

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

edited Nov 21, 2017 at 9:32

answered Nov 21, 2017 at 9:27

Youcef LAIDANI

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Comments

Glorfindel · Accepted Answer · 2017-11-21 09:38:44Z

0

I think this should work

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

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

edited Nov 21, 2017 at 9:38

Glorfindel

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answered Nov 21, 2017 at 9:38

Harshvinder Singh

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Comments

achAmháin · Accepted Answer · 2017-11-21 10:38:03Z

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!

Winter MC · Accepted Answer · 2020-07-17 17:48:10Z

0

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;
}

answered Jul 17, 2020 at 17:48

Winter MC

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