Regex for Hexadecimal characters

Question

I have written following program to write the IPV6 validation.

import java.util.regex.*;
import java.io.*;
import java.util.*;

public class demo
{
    //static String pattern = "([\\da-f]+):([\\da-f]+):([\\da-f]+):([\\da-f]+):([\\da-f]+):([\\da-f]+):([\\da-f]+):([\\da-f]+)";
    public static void main(String []args)
    {
        Pattern p=Pattern.compile(pattern);
        while(true)
        {
            Scanner sc = new Scanner(System.in);
            String line = sc.nextLine();
            if(line.equals(-1))
                System.exit(0);
            Matcher m =p.matcher(line);
            if(m.find())
                System.out.println("Found");
            else
                System.out.println("Not Found");
        }
}
}

But when I am running the code, it is printing "Found" for the below string. "1050:10:0:0:15:600:300c:326h"

I don't know why this is happening.(As last octet has 'h' in it, so it should print "NOT FOUND")

If you want to check if entire string matches regex you are looking for `matches` method. `find` checks if it can *find* any substring described by regex. If you can't change your method from `find` to `matches` then you need to surround your regex with `^` and `$` anchors. — Pshemo, Nov 22 '15 at 23:48
Does the ipv6 always has the same structure? If it is the case, you could add `{4}` for example instead of `+` for the last group — Yassin Hajaj, Nov 22 '15 at 23:48

Regex for Hexadecimal characters

0 Answers0