Java Regex Match and Replace matched group with characters of same length

Question

So I want to match credit card numbers and mask them in 6*4 format. So that only first 6 and last 4 characters will be visible. The characters between will be '*'. I tried to figure it out with a MASK like;

private static final String MASK = "$1***$3";
matcher.replaceAll(MASK);

But could not find out the way to give me back equal length of stars in the middle as the group $2.

Then I implemented the below code and it works. But what i want to ask if there is a shorter or easier way to do this. Anyone knows it?

private static final String HIDING_MASK = "**********";
private static final String REGEX = "\\b([0-9]{6})([0-9]{3,9})([0-9]{4})\\b";
private static final int groupToReplace = 2;

private String formatMessage(String message) throws NotMatchedException {
    Matcher m = Pattern.compile(REGEX).matcher(message);

    if (!m.find()) throw new NotMatchedException();
    else {
        StringBuilder maskedMessage = new StringBuilder(message);
        do {
            maskedMessage.replace(m.start(groupToReplace), m.end(groupToReplace), 
                    HIDING_MASK.substring(0, (m.end(groupToReplace) - m.start(groupToReplace))));

        } while(m.find(m.end()));

        return maskedMessage.toString();
    }
}

EDIT: Here is an example message to process. "2017.08.26 20:51 [Thread-Name] [Class-Name] [MethodName] Credit card holder 12345678901234567 02/2022 123 ........."

Why use a regex for this? `String masked = num.substring(0,6) + "******" + num.substring(12,16)` — slim, Aug 25 '17 at 16:01
If you really need a regex solution, then have a look at `Pattern#appendReplacement` — Wiktor Stribiżew, Aug 25 '17 at 16:04
@slim Because I don't know what is and how many credit card numbers(if there is) in the string — kerberos84, Aug 25 '17 at 16:04
Ah, hold on, are you talking about using this to find and replace numbers in a long string containing arbitrary text with credit card numbers within? If so you should edit the question to explain that -- give an example input text. — slim, Aug 25 '17 at 16:25

anomeric · Answer 1 · 2017-08-26T21:10:47.073

1

private String formatMessage(String message) throws NotMatchedException { 
    if (message.matches(".*\\b\\d{13,19}\\b.*")) {
        return message.replaceAll("(?:[.\\b]*)(?<=\\d{6})\\d(?=\\d{4})(?:[.\\b]*)", "*");
    } else {
        throw new NotMatchedException() ;
    }
}

edited Aug 26 '17 at 21:10

answered Aug 25 '17 at 16:02

anomeric

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I don't think this would be easy if the `message` contains the credit-card number only as a substring. – alirabiee Aug 25 '17 at 16:07
I am afraid you didn't understand my question. Credit card numbers are NOT always 16 digits. In my case I am looking for 13 to 19 digit cards. – kerberos84 Aug 25 '17 at 16:09
Thx but I already implemented a similar approach as you can see in my question. I am interested in doing this in a simpler way similar to ALI's answer. – kerberos84 Aug 25 '17 at 16:17
The replace all was also added to my answer – anomeric Aug 25 '17 at 16:22
hi your answer is still not a solution for this. I don't know what is in the 'message'. It is not consisting of a pure credit card number. An example message: "2017.08.26 20:51 [Thread-Name] [Class-Name] [MethodName] Credit card holder 12345678901234567 02/2022 123 ........." and so on. Hence I can't just check the length of the message as your answer suggest. But thx anyway. – kerberos84 Aug 26 '17 at 18:53
You can't expect an answer to part of a question you didn't include in your OP. I'll edit my answer – anomeric Aug 26 '17 at 18:55
hi, i thought it is self explanotary from the code i wrote. but anywat my fault so i added the example to the question. thanks. – kerberos84 Aug 26 '17 at 19:12
There were some typos in my last. These have been addressed – anomeric Aug 26 '17 at 19:19
hi @anomeric i get this exception; `Exception in thread "main" java.util.regex.PatternSyntaxException: Illegal/unsupported escape sequence near index 6 (?:[.\b]*)(?<=\d{6})\d(?=\d{4})(?:[.\b]*) ^ ` I changed the both `\\b` in the `replaceAll` mask to `\b` and now it does what i asked. Thanks for the answer. – kerberos84 Aug 27 '17 at 11:41

score 1 · Accepted Answer · answered Aug 25 '17 at 16:06

1

You can do it simply with this code:

str.replaceAll( "(?<=\\d{6})\\d(?=\\d{4})", "*" );

answered Aug 25 '17 at 16:06

alirabiee

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Does what OP asked. It's inefficient, but in exchange you get unreadability ;) – slim Aug 25 '17 at 16:12
It is close but it does not. My maximum card number is 19 digits.So can you edit it not to work if number is greater than 19 digits? – kerberos84 Aug 25 '17 at 16:15
1

`if(str.length() < 19)`? – slim Aug 25 '17 at 16:16
@kerberos84 Do a `if( str.matches( ".*\\b\\d{13,19}\\b.*" ) )` before – alirabiee Aug 25 '17 at 16:21
It still matches for "12345678901234567890" which is 20 digit? – kerberos84 Aug 25 '17 at 16:28
Not really, please make sure you copy the exact statement from the comment. – alirabiee Aug 25 '17 at 16:30
hi yes you are right. I mark this answer as acepted because this was written before the other answer. Thanks. – kerberos84 Aug 27 '17 at 12:01

linden2015 · Answer 3 · 2017-08-25T16:26:26.203

0

Readable but uncool.

String in = "1234561231234";
String mask = in
    .replaceFirst("^\\d{6}(\\d+)\\d{4}$", "$1")
    .replaceAll("\\d", "\\*");
String out = in
    .replaceFirst("^(\\d{6})\\d+(\\d{4})$", "$1" + mask + "$2");

edited Aug 25 '17 at 16:26

answered Aug 25 '17 at 16:15

linden2015

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score 0 · Answer 4 · answered Aug 25 '17 at 17:30

You can use the following if your text contains multiple credit-card numbers with variable lengths:

str.replaceAll( "\\b(\\d{13,19})\\b", "\u0000$1\u0000" )
   .replaceAll( "(?<=\\d{6})(?<=\u0000\\d{6,14})\\d(?=\\d{4,12}\u0000)(?=\\d{4})", "*" )
   .replaceAll( "\u0000([\\d*]+)\u0000", "$1" );

Not really readable, though, but it's all in one go.

slim · Answer 5 · 2017-08-25T16:15:16.597

A simple solution for a 16 char "number":

 String masked = num.substring(0,6) + "******" + num.substring(12,16);

For a string of arbitrary length ( >10 ):

 String masked = num.substring(0,6) 
               + stars(num.length() - 10) 
               + num.substring(num.length() - 6);

... where stars(int n) returns a String of n stars. See Simple way to repeat a String in java -- or if you don't mind a limit of 9 stars, "*********".substring(0,n)

score -1 · Answer 6 · answered Aug 25 '17 at 16:23

-1

Use a StringBuffer and overwrite the desired characters:

StringBuffer buf = new StringBuffer(num);
for(int i=4; i< buf.length() - 6) {
    buf.setCharAt(i, '*');
}
return buf.toString();

You could also use buf.replace(int start, int end, String str)

answered Aug 25 '17 at 16:23

slim

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Java Regex Match and Replace matched group with characters of same length

6 Answers6