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Java Word Count

I am just starting out in Java so I appreciate your patience. Anyways, I am writing a word count program as you can tell by the title, I am stuck at the numWords function below the for loop, I am not sure what I should set it equal to. If someone could set me in the right direction that would be awesome. Thank you. Here is all of my code thus far, let me know if I not specific enough in what I am asking, this is my first post. Thanks again.
import java.util.Scanner;
public class WCount {
public static void main (String[] args) {
Scanner stdin = new Scanner(System.in);
String [] wordArray = new String [10000];
int [] wordCount = new int [10000];
int numWords = 0;
while(stdin.hasNextLine()){
String s = stdin.nextLine();
String [] words = s.replaceAll("[^a-zA-Z ]", "").toLowerCase().split("\\s\
+");
for(int i = 0; i < words.length; i++){
numWords = 0;
}
}
}
}

If your code is intended to just count words, then you don't need to iterate through the words array at all. In other words, replace your for loop with just:
numWords += words.length;
Most likely a simpler approach would be to look for sequences of alpha characters:
Matcher wordMatch = Pattern.compile("\\w+").matcher();
while (wordMatch.find())
numWords++;
If you need to do something with the words (such as store them in a map to a count) then this approach will make that simpler:
Map<String,Integer> wordCount = new HashMap<>();
Matcher wordMatch = Pattern.compile("\\w+").matcher();
while (wordMatch.find()) {
String word = wordMatch.group();
int count = wordCount.getOrDefault(word, 0);
wordCount.put(word, count + 1);
}

Don't worry. We were all beginners once.
First of all, you don't need to do the loop because "length" attribute already has it. But, if you want to practice with loops is so easy as increasing the counter each time the iterator advances and that's it.
numWords++;

Hint: Read the input
String sentence = stdin.nextLine();
Split the string
String [] words = sentence.split(" ");
Number of words in a sentence
System.out.println("number of words in a sentence are " + words.length);
You mentioned in comments that you would also like to print the line in alphabetical order. For that Java got you covered:
Arrays.sort(words);

The best way to count the amount of words in a String String phrase is simply to get a String array from it using the String method split String[] words = phrase.split(" ") and giving it as argument the space itself, this will return a String array with each different words, then you can simple check its lengthwords.length and this will give you the exact number.

Take a input and add it to Linkedlist

How do i take a input from JOptionPane.showInputDialog, split it and add it to an Arraylist?
public static void main(String[] args) {
String acc = JOptionPane.showInputDialog("Enter a string:");
int num = Integer.parseInt(acc);
}

You can use Java's String.split() to separate a string based on a separatos.
For example if the words in the String are separated by a space then you can use:
yourString.split(" ");
This will return an String array. A more concrete example for what you want can be something like this
ArrayList<String> list = new ArrayList<String>();
String pop = "hello how are you doing";
for(String s: pop.split(" ")){
list.add(s);
}
The variable 'list' will contain:
["hello", "how", "are", "you", "doing" ]
EDIT: I read in another post that you wanted to parse it to integer first, you should put that kind of things in your question. If you can then it's better if you split the string with the above method and then parse each element as you add it to an Integer ArrayList (this if the elemenets are integers).

Please pay more attention to your posts. In the title you say LinkedList, in the text ArrayList. By the way will nobody figure out what you really want with that kind of information.
So you want to split something? You mean the string that you are getting there?
Then just look at this post!
How to split a string in Java
Then with the single values you simply add then to the list.
Example:
String acc = JOptionPane.showInputDialog("Enter a string:"); //enters yes-no
String[] result = acc.split("-");
myArrayList.add(result[0]); //yes
myArrayList.add(result[1]); //no

Component frame = new JFrame();
// get text
String name = JOptionPane.showInputDialog(frame, "What's your name?");
System.out.println(name);
// split text when u get a space
List<String> list = Arrays.asList(name.split("\\p{Z}+"));
System.out.println(list);

Beginner with 'nextLine'

I am working on a java problem at the moment where I am creating a program that simulates the old TV quiz show, You Bet Your Life. The game show host, Groucho Marx, chooses a secret word, then chats with the contestants for a while. If either contestant uses the secret word in a sentence, he or she wins $100.00.
My program is meant to check for this secret word.
Here is my code so far:
import java.util.Scanner;
public class Groucho {
String secret;
Groucho(String secret) {
this.secret = secret;
}
public String saysSecret(String line) {
if(secret.equals(line)){
return ("true");
} else {
return ("false");
}
}
public static void main(String[] args){
Scanner in = new Scanner(System.in);
}
}
In the main method I need to now create a new Groucho object with a secret word from the first line of standard input (in.nextLine()).
I am not sure how I go about doing this? Can someone explain please!
Thanks!
Miles

Have a look at the Scanner API, and perhaps the Java Tutorial on Objects. And that on Strings.
Learning the basics is usually more useful than just getting a line of code from somewhere.
No offence :).

You can read the line with the following statement:
String line = in.nextLine();
Then, if you'd like to have the first word (for example), you can split the line and create a new Groucho object.
String split = line.split(" ");
Groucho g = new Groucho(split[0]);
Here you can find more information about :
Scanner
String.split()

You would create a new Groucho object and pass in in.nextLine() as a parameter. This would be done by Groucho g = new Groucho( in.nextLine() );

You will need something that looks like this:
Scanner in = new Scanner(System.in); //take in word
String secretWord = in.nextLine(); //put it in a string
Groucho host = new Groucho (secretWord); //create a Groucho object and pass it the word

in.nextLine() will take a single line of the whole input, so you can simply pass it into the constructor.
For example:
String inputWord = in.nextLine();
Groucho g = new Groucho(inputWord);

In the Scanner class the nextLine() method takes the next line of input as a String. You can save that line of input to a String variable:
String line = in.nextLine();
Now that you have a full line of input, you can get the first word from it.
In a sentence each word is separated from other words by a space. In the String class the split() method can split a String into an array of smaller strings, such as words in a sentence, with a given separator, such as a space (" "), that you specify as a parameter:
String[] words = line.split(" ");
Next you can choose a secret word from the array by selecting the appropriate index.
For the first word:
String chosenWord = words[1];
For the last word:
String chosenWord = words[words.length - 1];
For a random word:
String chosenWord = words[Math.floor(Math.random() * words.length)];
Now you can simply pass on the secret word as a parameter to a new Groucho constructor:
Groucho secretWord = new Groucho(chosenWord);
This step by step explanation created a new variable at each step. You can accomplish the same task by combining multiple lines of code into a single statement and avoid creating unnecessary variables.

java store a string in a char array

In Java, How can I store a string in an array? For example:
//pseudocode:
name = ayo
string index [1] = a
string index [2] = y
string index [3] = o
Then how can I get the length of the string?
// this code doesn't work
String[] timestamp = new String[40]; String name;
System.out.println("Pls enter a name and surname");
Scanner sc = new Scanner(System.in);
name = sc.nextLine();
name=timestamp.substring(0, 20);

If you want a char array to hold each character of the string at every (or almost every index), then you could do this:
char[] tmp = new char[length];
for (int i = 0; i < length; i++) {
tmp[i] = name.charAt(i);
}
Where length is from 0 to name.length.

This code doesn't compile because the substring method can only be called on a String, not a String array if I'm not mistaken. In the code above, timestamp is declared as a String array with 40 indexes.
Also in this code, you're asking for input from a user and assigning it to name in this line:
name = sc.nextLine();
and then you are trying to replace what the user just typed with what is stored in timestamp on the next line which is nothing, and would erase whatever was stored in name:
name = timestamp.substring(0,20);
And again that wouldn't work anyway because timestamp is an array of 40 strings instead of one specific string. In order to call substring it has to be just one specific string.
I know that probably doesn't help much with what you're trying to do, but hopefully it helps you understand why this isn't working.
If you can reply with what you're trying to do with a specific example I can help direct you further. For example, let's say you wanted a user to type their name, "John Smith" and then you wanted to seperate that into a first and last name in two different String variables or a String array. The more specific you can be with what you want to do the better. Good luck :)
BEGIN EDIT
Ok here are a few things you might want to try if I understand what you're doing correctly.
//Since each index will only be holding one character,
//it makes sense to use char array instead of a string array.
//This next line creates a char array with 40 empty indexes.
char[] timestamp = new char[40];
//The variable name to store user input as a string.
String name;
//message to user to input name
System.out.println("Pls enter a name and surname");
//Create a scanner to get input from keyboard, and store user input in the name variable
Scanner sc = new Scanner(System.in);
name = sc.nextLine();
//if you wanted the length of the char array to be the same
//as the characters in name, it would make more sense to declare it here like this
//instead of declaring it above.
char[] timestamp = new char[name.length()];
//For loop, loops through each character in the string and stores in
//indexes of timestamp char array.
for(int i=0; i<name.length;i++)
{
timestamp[i] = name.charAt(i);
}
The other thing you could do if you wanted to just seperate the first and last name would be to split it like this.
String[] seperateName = name.split(" ");
That line will split the string when it finds a space and put it in the index in the seperateName array. So if name was "John Smith", sperateName[0] = John and seperateName[1] = Smith.

Are you looking for a char[]? You can convert a character array to a String using String.copyValueOf(char[]).

Java, substring an array:
Use Arrays.copyOfRange:
public static <T> T[] copyOfRange(T[] original,
int from,
int to)
For example:
import java.util.*;
public class Main{
public static void main(String args[]){
String[] words = new String[3];
words[0] = "rico";
words[1] = "skipper";
words[2] = "kowalski";
for(String word : words){
System.out.println(word);
}
System.out.println("---");
words = Arrays.copyOfRange(words, 1, words.length);
for(String word : words){
System.out.println(word);
}
}
}
Prints:
rico
skipper
kowalski
---
skipper
kowalski
Another stackoverflow post going into more details:
https://stackoverflow.com/a/6597591/445131

Algorithm to output the initials of a name

I have just started the java programming and at the moment I am doing the basic things. I came across a problem that I can't solve and didn't found any answers around so I thought you might give me a hand. I want to write a program to prompt the user to enter their full name (first name, second name and surname) and output their initials.
Assuming that the user always types three names and does not include any unnecessary spaces. So the input data will always look like this : Name Middlename Surname
Some of my code that I have done and stuck in there as I get number of the letter that is in the code instead of letter itself.
import java.util.*;
public class Initials
{
public static void main (String[] args)
{
//create Scanner to read in data
Scanner myKeyboard = new Scanner(System.in);
//prompt user for input – use print to leave cursor on line
System.out.print("Please enter Your full Name , Middle name And Surname: ");
String name = myKeyboard.nextLine();
String initials1 = name.substring(0, 1);
int initials2 = name.
//output Initials
System.out.println ("Initials Are " + initials1 + initials2 + initials3);
}
}

Users will enter a string like
"first middle last"
so therefore you need to get each word from the string.
Loot at split.
After you get each word of the user-entered data, you need to use a loop to get the first letter of each part of the name.

First, the nextLine Function will return the full name. First, you need to .split() the string name on a space, perhaps. This requires a correctly formatted string from the user, but I wouldn't worry about that yet.
Once you split the string, it returns an array of strings. If the user put them in correectly, you can do a for loop on the array.
StringBuilder builder = new StringBuilder(3);
for(int i = 0; i < splitStringArray.length; i++)
{
builder.append(splitStringArray[i].substring(0,1));
}
System.out.println("Initials Are " + builder.toString());

Use the String split() method. This allows you to split a String using a certain regex (for example, spliting a String by the space character). The returned value is an array holding each of the split values. See the documentation for the method.
Scanner myKeyboard = new Scanner(System.in);
System.out.print("Please enter Your full Name , Middle name And Surname: ");
String name = myKeyboard.nextLine();
String[] nameParts = name.split(" ");
char firstInitial = nameParts[0].charAt(0);
char middleInitial = nameParts[1].charAt(0);
char lastInitial = nameParts[2].charAt(0);
System.out.println ("Initials Are " + firstInitial + middleInitial + lastInitial);
Note that the above assumes the user has entered the right number of names. You'll need to do some catching or checking if you need to safeguard against the users doing "weird" things.