CODEWITHSUNDEEP
javaregexstringreplaceall
Cassie
Cassie

Reputation: 3099

ReplaceAll with defined replacement number Java

I have a method for email masking. I need to replace letters in the email before @ sign with stars. But the problem is that there is always should be exactly 5 stars and the first and last elements should not be hidden. A sample input would be: [email protected]. Output: s*****[email protected] So it does not matter how many characters between the first and the last one in the e-mail. Here is my code:

public static String maskEmail(String inputEmail){
    return inputEmail.replaceAll("(?<=.).(?=[^@]*?.@)", "*");
}

My method masks this e-mail, but the problem is that I don't know how to put 5 stars exactly.

Upvotes: 2

Views: 51

Answers (3)

shmosel
shmosel

Reputation: 50716

How about this:

inputEmail.replaceAll("(?<=^.).*(?=.@)", "*****")

Or this:

inputEmail.replaceAll("(.).*(.@)", "$1*****$2")

Note that this only works if there are at least 2 characters before the @.

Upvotes: 3

anto
anto

Reputation: 21

Try this code:

import java.util.Arrays;

public class HelloWorld{

    public static String StrToAsterisk(String email){
        if (email == null) return "";

        int flag = email.indexOf("@");
        if (flag < 0) return "";

        StringBuilder sb = new StringBuilder();
        sb.append(email.charAt(0));
        sb.append("*****");
        sb.append(email.substring(flag-1));
        return sb.toString();
     }

     public static void main(String []args){
         System.out.println(StrToAsterisk("[email protected]"));
         //input : [email protected]
         //output: s*****[email protected]
     }
 }

Upvotes: 2

Mureinik
Mureinik

Reputation: 311418

It would be much simpler to just take the first letter and concatenate it with five asterisks and the substring starting from the letter before the @:

public static String maskEmail(String inputEmail) {
    return inputEmail.substring(0, 1) + 
           "*****" + 
           inputEmail.substring(inputEmail.indexOf('@') - 1);
}

Upvotes: 5

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