masking of email address in java

Question

I am trying to mask email address with "*" but I am bad at regex.

input : nileshxyzae@gmail.com
output : nil********@gmail.com

My code is

String maskedEmail = email.replaceAll("(?<=.{3}).(?=[^@]*?.@)", "*");

but its giving me output nil*******e@gmail.com I am not getting whats getting wrong here. Why last character is not converted? Also can someone explain meaning all these regex

Answer 1

Your look-ahead (?=[^@]*?.@) requires at least 1 character to be there in front of @ (see the dot before @).

If you remove it, you will get all the expected symbols replaced:

(?<=.{3}).(?=[^@]*?@)

Here is the regex demo (replace with *).

However, the regex is not a proper regex for the task. You need a regex that will match each character after the first 3 characters up to the first @:

(^[^@]{3}|(?!^)\G)[^@]

See another regex demo, replace with $1*. Here, [^@] matches any character that is not @, so we do not match addresses like abc@example.com. Only those emails will be masked that have 4+ characters in the username part.

See IDEONE demo:

String s = "nileshkemse@gmail.com";
System.out.println(s.replaceAll("(^[^@]{3}|(?!^)\\G)[^@]", "$1*"));

Answer 2

If you're bad at regular expressions, don't use them :) I don't know if you've ever heard the quote:

Some people, when confronted with a problem, think "I know, I'll use regular expressions." Now they have two problems.

(source)

You might get a working regular expression here, but will you understand it today? tomorrow? in six months' time? And will your colleagues?

An easy alternative is using a StringBuilder, and I'd argue that it's a lot more straightforward to understand what is going on here:

StringBuilder sb = new StringBuilder(email);
for (int i = 3; i < sb.length() && sb.charAt(i) != '@'; ++i) {
  sb.setCharAt(i, '*');
}
email = sb.toString();

"Starting at the third character, replace the characters with a * until you reach the end of the string or @."

(You don't even need to use StringBuilder: you could simply manipulate the elements of email.toCharArray(), then construct a new string at the end).

Of course, this doesn't work correctly for email addresses where the local part is shorter than 3 characters - it would actually then mask the domain.

Answer 3

Your Look-ahead is kind of complicated. Try this code :

public static void main(String... args) throws Exception {
    String s = "nileshkemse@gmail.com";
    s= s.replaceAll("(?<=.{3}).(?=.*@)", "*");
    System.out.println(s);
}

O/P :

nil********@gmail.com

Answer 4

//In Kotlin

val email = "nileshkemse@gmail.com"
val maskedEmail = email.replace(Regex("(?<=.{3}).(?=.*@)"), "*")

Answer 5

I like this one because I just want to hide 4 characters, it also dynamically decrease the hidden chars to 2 if the email address is too short:

public static String maskEmailAddress(final String email) {
    final String mask = "*****";
    final int at = email.indexOf("@");
    if (at > 2) {
        final int maskLen = Math.min(Math.max(at / 2, 2), 4);
        final int start = (at - maskLen) / 2;
        return email.substring(0, start) + mask.substring(0, maskLen) + email.substring(start + maskLen);
    }
    return email;
}

Sample outputs:

my.email@gmail.com    >    my****il@gmail.com
info@mail.com         >    i**o@mail.com

Answer 6

    public static string GetMaskedEmail(string emailAddress)
    {
        string _emailToMask = emailAddress;
        try
        {
            if (!string.IsNullOrEmpty(emailAddress))
            {
                var _splitEmail = emailAddress.Split(Char.Parse("@"));
                var _user = _splitEmail[0];
                var _domain = _splitEmail[1];

                if (_user.Length > 3)
                {
                    var _maskedUser = _user.Substring(0, 3) + new String(Char.Parse("*"), _user.Length - 3);
                    _emailToMask = _maskedUser + "@" + _domain;
                }
                else
                {
                    _emailToMask = new String(Char.Parse("*"), _user.Length) + "@" + _domain;
                }
            }
        }
        catch (Exception) { }
        return _emailToMask;
    }

Answer 7

Kotlin Extension

/**
 * Format email id in the following format
 * (m••••••n@gmail.com)
 * only exception for 2 character username in that case it will be ••@gmail.com
 * @return formatted email
 */
fun String.maskEmailId(): String {
    return if (this.isNotNullOrEmpty() && this.indexOf('@') > 0) {
        val index = this.indexOf('@')
        val maskedUsername = if (index > 2) {
            "${this.substring(0, 1)}${"*".repeat(index-2)}${this.substring(index-1)}"
        } else {
            "*".repeat(index) + this.substring(index)
        }
        maskedUsername
    }  else ""
}