$10 for answer: using Pattern - regular expressions and Scanner

Question

I'm trying to read input messages using Scanner. The messages are in format: word1 word2 number. the "number" is followed by "C null terminator" i.e. '\u0000' character.
I tried this:
Scanner input;
          Pattern pat = Pattern.compile("(\\w+) (\\w+) (\\d+)\u0000");
          input.useDelimiter(pat);
          while (input.hasNext(pat)) {
The input is there and in correct format, but it hangs hasNext(pat) - as if input is not correct. What did I do wrong? Thanks in advance - thsi is urgent, I'd be glad to send you $10 for correct answer via PayPal.
Message was edited by:
MrM654

Answer 1

Here is a Regex and a Scanner implementation.
The price is $20 because there are 2 ways to do this.
Plus at least a $10 courtesy tip.
import java.util.Scanner;
import java.util.regex.*;
public class RegexTest{
public static void main(String[] args){
     new RegexTest();
public RegexTest(){
     String input = "word1 word1 1\u0000 word2 word2 2\u0000 word3 word3 3\u0000";
     Pattern pattern = Pattern.compile("(\\w+ \\w+ \\d+\\u0000)");
     Matcher matcher = pattern.matcher(input);
     while(matcher.find() == true){
     System.out.println("Match: " + matcher.group());
     String input = "word word 1\u0000 word2 word2 2\u0000 word3 word3 3\u0000";
     Pattern pattern = Pattern.compile("(\\w+) (\\w+) (\\d+)\\u0000\\s*");
    Scanner scanner = new Scanner(input);
     String result;
     while((result = scanner.findInLine(pattern)) != null){
     System.out.println("Match: " + result);
     System.out.println("Done.");
}Edit: Changed Pattern to end with \\s* to be insensitive to ending spaces