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
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;
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")
10\\..*. But could you take a look on regex syntax? It is not a big deal, believe me. People will be much more motivated to answer your questions if you can show some efforts you performed before asking the question.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?