RegEx to validate ip address with wildcard (last index only) in Java

Question

I can't seem to get the proper RegEx for validating an IP address, including support for a wildcard char (*), which can occur only at the end (last index) means * (asterisk) can only available after 3rd '.'(dot) . For example:

Valid IP

• 0.0.0.*
• 255.255.255.*

Invalid IP

• 0.*
• 255.*
• 256.*
• 0.0.*
• 255.255.*
• 256.256.*

Answer 1

String regex = "^((0|255)\\.){3}([0-9]|[1-9][0-9]|[1-2][0-5][0-5])$";

Answer 2

You can use below code for identifying IP address

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class IPAddressValidator{

    private Pattern pattern;
    private Matcher matcher;

    private static final String IPADDRESS_PATTERN =
        "^([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])\\." +
        "([01]?\\d\\d?|2[0-4]\\d|25[0-5])$";

    public IPAddressValidator(){
      pattern = Pattern.compile(IPADDRESS_PATTERN);
    }

   /**
    * Validate ip address with regular expression
    * @param ip ip address for validation
    * @return true valid ip address, false invalid ip address
    */
    public boolean validate(final String ip){
      matcher = pattern.matcher(ip);
      return matcher.matches();
    }
}

Could change Regex according to your requirement

Answer 3

Something like this should do it:

^(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|\*)$

Quite accurate i think for this purpose