Regex with exception

Question

i need regex for detect all IPs with ports and without ports but excepting

93.153.31.151(:27002)

and

10.0.0.1(:27002)

I have got some but I need add exception

\\d{1,3}(?:\\.\\d{1,3}){3}(?::\\d{1,5})?

For java matcher

    String numIPRegex = "\\d{1,3}(?:\\.\\d{1,3}){3}(?::\\d{1,5})?";

    if (pA.matcher(msgText).find()) {
        this.logger.info("Found");

    } else {    
        this.logger.info("Not Found");                  

    }

Regexes don't seem like the right choice here. You have InetAddress which can detect correctly formatted addresses, Integer.parseInt() which can parse integers, and indexOf() in String. And, if you use Guava, you also have HostAndPort. — fge
– fge, Commented Jan 6, 2013 at 13:27
@JanDvorak good one, I didn't know about it! Looks like InetSocketAddress.createUnresolved() is even better (especially since InetAddress may attempt name resolution). — fge
– fge, Commented Jan 6, 2013 at 13:41

Tomalak · Accepted Answer · 2013-01-06 14:16:35Z

4

Without making a statement about better-suited Java classes that can handle IP addreses in a structured manner...

You can add exceptions to a regular expression using negative look-aheads:

String numIPRegex = "(?!(?:93\\.153\\.31\\.151|10\\.0\\.0\\.1)(?::27002)?)\\d{1,3}(?:\\.\\d{1,3}){3}(?::\\d{1,5})?";

Explanation:

(?!                                   # start negative look-ahead
  (?:                                 #   start non-capturing group
    93\.153\.31\.151                  #     exception address #1
    |                                 #     or
    10\.0\.0\.1                       #     exception address #2
  )                                   #   end non-capturing group
  (?:                                 #   start non-capturing group
    :27002                            #     port number
  )?                                  #   end non-capturing group; optional
)                                     # end negative look-ahead
\d{1,3}(?:\.\d{1,3}){3}(?::\d{1,5})?  # your original expression

Of course the other obvious alternative would be to test the exceptions up-front one by one and just return false if one exceptions matches. Wrapping them up all up in one single big regex will quickly become very ugly.

edited Jan 6, 2013 at 14:16

answered Jan 6, 2013 at 13:37

Tomalak

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1 Comment

Ross Patterson Over a year ago

+1 for the implied "You shouldn't use regular expressions for something like this" and for supplying an educational answer anyway.

ggrandes · Accepted Answer · 2013-01-06 13:56:52Z

0

You want this?

public static void main(String[] args) {
    System.out.println(match("93.153.31.151(:27002)")); // false
    System.out.println(match("12.23.34.45(:21002)")); // true
}
public static boolean match(String input) {
    String exclude = "|93.153.31.151(:27002)|10.0.0.1(:27002)|";
    if (exclude.contains("|" + input + "|")) return false; // Exclude from match
    //
    String numIPRegex = "^\\d{1,3}(\\.\\d{1,3}){3}\\(:\\d{1,5}\\)$";
    Pattern pA = Pattern.compile(numIPRegex);
    //
    if (pA.matcher(input).matches()) {
        System.out.println("Found");
        return true;
    } else {    
        System.out.println("Not Found");                  
    }
    return false;
}

answered Jan 6, 2013 at 13:56

ggrandes

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