使用java.util.regex.Pattern在java中查找类似的IP

Question

I am trying to find similar IP using java.util.regex.Pattern in java.

String ipAddr = "192.168.112.33";
Pattern PRIVATE_ADDRESS_PATTERN = Pattern.compile("192.168.(.*?).33", Pattern.CASE_INSENSITIVE);
String IPaddress = PRIVATE_ADDRESS_PATTERN.matcher(ipAddr).toString(); 

It is not working for me. Where is my mistake?

Answer 1

Why you simply don't make :

String ipAddr = "192.168.112.33";

//If your pattern match with your String then it is correct else it is not
if (ipAddr.matches("192\\.168\\.(.*?)\\.33")) {
    System.out.println("CORRECT");
} else {
    System.out.println("NOT CORRECT");
}

Like @lathspell said in the comment you have to escape the dot (.) with \\\\.

Answer 2

A more correct matcher would be the following:

Pattern IP_PATTERN = Pattern.compile("192\\.168\\.(\\d{1,3})\\.33");

This way you would only match numbers in there and not letters, for instance. According to further use cases, you may want to refine your number matcher to make sure it matches only in the range 0-255 :

String IP_PATTERN_PART = "(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9]?)";
Pattern IP_PATTERN = Pattern.compile("192\\.168\\.("+IP_PATTERN_PART+")\\.33");

Also, if you are sure you will never get anything else than an IP (meaning no enclosing text), you may want to use the start and end delimiters:

Pattern IP_PATTERN = Pattern.compile("^192\\.168\\.(\\d{1,3})\\.33$");

Then, once you've decided which pattern to use, you'll have to use it this way:

Matcher matcher = IP_PATTERN.matcher(candidate);
String otherIP = null;
if (matcher.matches()) {
  matcher.find();
  otherIP = matcher.group(1);
}