Java regex pattern matching length

Question

I have input string as

input = "AAA10.50.30.20"

input.replaceAll("10.([01]?\\d\\d?|2[0-4]\\d|25[0-5]).([01]?\\d\\d?|2[0-4]\\d|25[0-5]).([01]?\\d\\d?|2[0-4]\\d|25[0-5])",
                        "X");

output is : "AAAX"

However i want the output as "AAAXXXXXXXXXXX"

It should replace the IP with multiple 'X' which are equivalent to number of characters in IP address

Answer 1

If you just want to replace all digits and dots by X call str.replaceAll("\\\\d|\\\\.", "X") . If you want to match the exact pattern use something like

str.replaceAll("(?:\\d{1,3}\\.){3}\\d{1,3}", "X")

Answer 2

String input =  "AAA10.50.30.20";
Pattern p = p = Pattern.compile("10.([01]?\\d\\d?|2[0-4]\\d|25[0-5]).([01]?\\d\\d?|2[0-4]\\d|25[0-5]).([01]?\\d\\d?|2[0-4]\\d|25[0-5])");
Matcher m = p.matcher(input);
String result = input;
while(m.find()){
    char[] replacement = new char[m.end()-m.start()];
    Arrays.fill(replacement, 'X');
    result = result.substring(0, m.start())
        + new String(replacement)
        + result.substring(m.end());
}
return result;

Answer 3

The problem is that you search for the IP-address with your regex. That gives you 1 match - the IP-address. You then replace it with X, giving you AAAX. How about fetching the string matched by the regex instead, then replacing all digits with X?