New to Java

I need to know what I'm doing wrong here..

Here is my objective:

Write a program to read a sentence (a String) entered by the user and display the following statistics:
1. the number of words in the sentence;
2. the length of each word (the number of characters);
3. the length of the longest word; and
4. the length of the shortest word
5. assume that the phrase will always end with punctuation and the punctuation will not be counted as a character in the computation of the length.

example:


Please enter the sentence to analyse:

Assignment2 may require a loop!

Here are some statistics on your sentence:
Word 1 has 11 characters.
Word 2 has 3 characters.
Word 3 has 7 characters.
Word 4 has 1 character.
Word 5 has 4 characters.
There are 5 words.
The longest word has 11 characters.
The shortest word has 1 character.
Press any key to continue...



Now I was able to code enough to find the amount of words in a string, but I'm stuck when I try to compute the length of each word :(

If anybody can help, it would be greatly appreciated

Oh and I CANNOT use StringTokenizer :(


My code:

import java.util.Scanner;

public class Assignment_2 
{

    public static void main (String args[]) 
    	
    {   
    	
        Scanner userInput = new Scanner(System.in);
    	
    	String a = "";
    	
    	a = userInput.nextLine();
    	
	  int textLength = a.length();
	  int numberOfWords = 0;
	  int countChar = 0;
	  boolean isWord = false;

		for (int i=0;i<textLength;i++)
		{
		   if (a.charAt(i) == ' ' || a.charAt(i) == '.'|| a.charAt(i) == '?' || a.charAt(i) == '!')
		   {      
			  numberOfWords++; \\when a space or a punctuation is found, increment number of words by one
			  isWord = true; \\ flag isWord as true when ending of a word is found
		   }
			
			else 			
			{	
			  countChar++; \\ otherwise we are inside a word and we need to increment the character count
			
			}
			
		    \\ I tried to put a while (isTrue) statement here, but it just  loops on forever.
			
		}
			System.out.println("Word " + numberOfWords); 
    	}
    }