[staykovmarin]

If you want all the newest features, then you should definitely get jdk6. I use 5, because there is nothing in 6 that i need.

You can use a Regex to do that:

Code:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RegexFinder {
	public static void main(String args[]) {
		try {
			new RegexFinder();
		} catch (IOException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
	}

	RegexFinder() throws IOException {
		// this is the Regex pattern:
		// any vowels: aeiou
		// spaces: \\s
		// punctioation : ?.,! (note: the ! must be escaped \\! because in Regex
		// it means NOT. we are just looking for the ! char
		Pattern pattern = Pattern.compile("[aeiou\\s?.,\\!]");
		// edit: forgot that you needed it to read from command line
		while (true) {
			System.out.print("Enter the string to search: ");
			BufferedReader in = new BufferedReader(new InputStreamReader(
					System.in));
			Matcher m = pattern.matcher(in.readLine());

			// all the counter variables
			int a = 0, e = 0, i = 0, o = 0, u = 0, space = 0, punks = 0;
			while (m.find()) {
				// this switch looks trough each character matched by the regex,
				// and
				// evaluates it
				switch ((int) m.group().charAt(0)) {
				case 'a':
					a++;
					break;
				case 'e':
					e++;
					break;
				case 'i':
					i++;
					break;
				case 'o':
					o++;
					break;
				case 'u':
					u++;
					break;
				case ' ':
					space++;
					break;
				default:
					// if it is not any of the above, but still found by the
					// regex,
					// it must be a punctioation mark
					punks++;
				}
			}

		System.out.println("# of a=" + a + "\n# of e=" + e + "\n# of i=" + i
				+ "\n# of o=" + o + "\n# of u=" + u + "\n# of spaces=" + space
				+ "\n# of punctuation marks=" + punks);
		}
	}
}