Develop Reference

While-loop will not terminate in console after entering numbers

I keep trying to get this to work but when I enter in the numbers and enter them into the console it does not finish. I have to terminate myself.
import java.util.Scanner;
public static void main(String[] args) {
int cmlSum = 0;
int inputNum;
String outputSum = "";
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter sequence of numbers ");
do {
inputNum = keyboard.nextInt();
cmlSum += inputNum;
outputSum += String.format("%s ", String.valueOf(cmlSum));
} while (keyboard.hasNextInt());
System.out.println(outputSum);
}

Well, yes. The keyboard.hasNextInt() call will return false for two reasons.
The next token is a NOT an integer.
You have reached the end-of-input.
What is (most likely) happening is that you have stopped entering numbers. The program is (patiently) waiting for you to enter ... something.
Solutions:
Tell the user to enter the (OS specific) terminal "end of file" character. On Linux it is CTRL-D. On Windows CTRL-Z.
Tell the user to enter something that isn't an integer.
Pick an integer as meaning that there are no more numbers, and test for that.
You also need to instruct the user how to "end" the sequence; e.g.
System.out.println("Enter sequence of numbers. Enter a non-number to stop.");
This is actually a problem with your application's "user interface" design. If the user is expected to type an arbitrarily long sequence of numbers (or something else), then there needs to be some way for the user to tell the program that the sequence is finished. The program cannot magically distinguish the cases of "there are no more" and "hang on, I'm taking a break from typing".

The hasNext() method checks if the Scanner has another token in its input. A Scanner breaks its input into tokens using a delimiter pattern, which matches whitespace by default. That is, hasNext() checks the input and returns true if it has another non-whitespace character.
In this case hasNext() won't return true because there is neither any integer nor any whitespace. Therefore the program waits for the next input. Besides use a specific integer to break the loop.
for instance,
System.out.println("Input -1 will end the program!";
do{
int x = keyboard.nextInt();
if(x == -1){
break;
}
//do something
}while(true);

Your code is ok. There is no issue.
But before writing code, we need to think about it. The workflow of your code below:
1st time when we enter do loop, keyboard.nextInt() takes input from us.
Then it calculates the sum and performs string operation.
After that, while's keyboard.hasNextInt() takes next input from you.
Checks your input. If your input is not an integer, while loop will terminate(break).
If your input is an integer then, code loop back to keyboard.nextInt(). But this time, it does not take input from you.
It pases the buffered input(keyboard.hasNextInt()) to keyboard.nextInt() and assign the value to inputNum
So, when you want to terminate while loop, you should input any character like a, b, c, etc.

You haven't specified when the loop will end. Have a condition such as inputting a certain number that will end the program once entered, as currently your program is just going to wait for more input. Something like :
System.out.println("Enter sequence of numbers to add. Enter '0' to end the program");
do {
inputNum = keyboard.nextInt();
cmlSum += inputNum;
outputSum += String.format("%s ", String.valueOf(cmlSum));
} while (inputNum != 0);//Keeps going as long as 0 is not entered
//When zero is entered, program shows the total sum and terminates
if (inputNum == 0) {
System.out.println("The sum of all total numbers: ");
System.out.println(outputSum);
System.exit(0);//Terminates program
}

Basic syntax of do-while Loop:
do{
// do something
}while(terminating condition);
If you are using hasNextInt() method of Scanner object for terminating condition in do-while loop then loop will be terminated once it get input other than an integer value (e.g float, double, char, String etc.. ) as shown in below complete program.
import java.util.Scanner;
public class Cumulative{
public static void main(String[] args){
int cmlSum = 0;
int inputNum;
String outputSum = "";
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter sequence of numbers ");
do{
inputNum = keyboard.nextInt();
cmlSum += inputNum;
outputSum += String.format("%s ", String.valueOf(cmlSum));
}while (keyboard.hasNextInt()); // loop will terminated whenever get any value other than valid integer such as float char or String etc..
System.out.println(outputSum);
}
}

Two problems, using charAt for undefined input and looping output

So, I posted this nearly identical code yesterday, asking about how to leave the punctuation at the end of a reversed sentence after using .split. I'm still struggling with it, but I'm also having another issue with the same code: And here is my screen shot http://i.stack.imgur.com/peiEA.png
import java.util.Scanner;
import java.util.StringTokenizer; // for splitting
public class MyTokenTester
{
public static void main(String\[\] args)
{
Scanner enter = new Scanner(System.in);
String sentinel = ""; // condition for do...while
String backward = ""; // empty string
char lastChar = '\0';
do
{
System.out.println("Please enter a sentence: ");
String sentence = enter.nextLine();
String\[\] words = sentence.split(" "); // array words gets tokens
// System.out.printf("The string is%s",sentence.substring(sentence.length()));
for (int count = words.length -1; count>=0; count--) // reverse the order and assign backward each token
{
backward += words\[count\] + " ";
}
System.out.println(backward); // print original sentence in reverse order
System.out.println("Hit any key to continue or type 'quit' to stop now: ");
sentinel = enter.nextLine();
sentinel = sentinel.toLowerCase(); // regardless of case
} while (!sentinel.equals("quit")); // while the sentinel value does not equal quit, continue loop
System.out.println("Programmed by ----");
} // end main
} // end class MyTokenTester][1]][1]
As you guys can probably see my from screen shot, when the user is prompted to add another sentence in, the previous sentence is read back again.
My questions are:
How do I use charAt to identify a character at an undefined index (user input with varying lengths)
How do I stop my sentence from reading back after the user decides to continue.
Again, as I said, I'd posted this code yesterday, but the thread died and I had additional issues which weren't mentioned in the original post.

To address part 2, if you want to stop the sentence from reading back previous input, then reset backward to an empty string, because as it stands now, you're constantly adding new words to the variable. So to fix this, add this line of code right before the end of your do-while loop,
backward = "";
To address part 1, if you want to check the last character in a string, then first you have to know what is the last index of this string. Well, a string has indexes from 0 to str.length()-1. So if you want to access the very last character in the user input, simply access the last word in your words array (indexed from 0 to words.length - 1) by doing the following,
words[count].charAt(words[count].length() - 1);
Note that count is simply words.length - 1 so this can be changed to your liking.

1) So you have this array of strings words. Before adding each word to the backward string, you can use something like: words[count].chartAt(words[count].length() - 1). It will return you the charater at the last position of this word. Now you are able to do you checking to know wether it is a letter or any special char.
2) The problem is not that it is reading the previous line again, the problem is that the backward string still has the previous result. As you are using a + operator to set the values of the string, it will keep adding it together with the previous result. You should clean it before processing the other input to have the result that you want.

here is your code:
import java.util.*;
public class main{
public static void main(String[] args){
Scanner enter = new Scanner(System.in);
String sentinel = ""; // condition for do...while
String backward = ""; // empty string
char lastChar = '\0';
do
{
System.out.println("Please enter a sentence: ");
String sentence = enter.nextLine();
String[] words = sentence.split(" "); // array words gets tokens
// System.out.printf("The string is%s",sentence.substring(sentence.length()));
List<String> items = Arrays.asList(words);
Collections.reverse(items);
System.out.println(generateBackWardResult(items)); // print original sentence in reverse order
System.out.println("Hit any key to continue or type 'quit' to stop now: ");
sentinel = enter.nextLine();
// i use quals ignore case, makes the code more readable
} while (!sentinel.equalsIgnoreCase("quit")); // while the sentinel value does not equal quit, continue loop
System.out.println("Programmed by ----");
} // end main
static String generateBackWardResult(List<String> input){
String result="";
for (String word:input){
result =result +" "+word;
}
return result;
}
} // end class MyTokenTester][1]][1]
there are also some thing to mention:
* never invent the wheel again! (for reverting an array there are lots of approaches in java util packages, use them.)
*write clean code, do each functionality, i a separate method. in your case you are doing the reverting and showing the result in a single method.

Keep on prompting user after invalid input

Now I know that there is a thread called "Validating input using java.util.Scanner". I already looked there and that thread only answered 1/2 of my problems. The other half is when someone enters a number greater than 2 I get Array Index Out of Bounds Exception. I just need help on if someone enters a 3 for either row or column, the console should prompt something like this:
"Enter the coordinates to place an 'X'. Row then Column."
//enters 3 and 3
"Please enter a valid input"
It would keep and asking the user for a valid number until he gives one.
Would I need to do something like the !keyboard.hasNextInt() but for integers? And that would run smoothly with the rest of my code?

You could use a do-while loop. Something like
do {
//prompt
//input
} while (input not valid);
Where prompt and input should be replaced by code to prompt the user and accept input. In the while section, check if input is valid.

You're question isn't too clear but I'll try to make sense of it.
I'm assuming you've named your scanner "keyboard"
Before I try running this code, the first problem I can see is this (Note that I grabbed this from your code before you edited the question):
while (board[row][col] != ' ')
{
System.out.println("Already occupied space");
System.out.println("Choose again");
row = keyboard.nextInt();
col = keyboard.nextInt();
}
Earlier, you made sure that the user enters integers. However, you have abandoned that completely in this case.
Assuming you're trying to avoid an error if the user enters something other than an integer, this is what I would do:
while(true){
boolean valid = true;
if(!keyboard.hasNextInt()){
valid = false;
keyboard.next();
}
else{
row = keyboard.nextInt();
}
if(!keyboard.hasNextInt()){
valid = false;
keyboard.next();
}
else{
col = keyboard.nextInt();
}
if (valid && (row > 2 || col > 2)){
System.out.println("Please enter a valid input");
continue;
}
else if(!valid){
System.out.println("Please enter a valid input");
continue;
}
else
break;
}
There are a couple reasons this code might seem a bit long. First off, we're trying to test if the input is an integer before we attempt to store it as an int. Secondly, we want to compare the input after we store it successfully to see if it's less than 3. If the input isn't an integer, the boolean "valid" will be false. The way a compiler works, if valid is false in the if statement it will ignore anything to the right of the &&, avoiding an error.
I admit, this is using some commands that I haven't learned before, so this might not be the most efficient way. But you get the idea :)
P.S. You should probably throw the above code into a method.

I want to know how to find a special character in a string

I am programming in Java in Eclipse, I want to ask users to enter their specific ID, Which starts with an uppercase G and has 8 digit numbers. like G34466567. if the user enters an invalid ID it will be an error. how can i separate a valid ID from others?

You can use regex. This pattern checks if the first character is a capital G and there are 8 digits following:
([G]{1})([0-9]{8})$
As you see there are two expressions which are separated by the (). The first says "only one character and this one has to be a capital G". And the second one says, there have to be 8 digits following and the digits can be from 0 to 9.
Every condition contains two "parts". The first with the [] defines which chars are allowed. The pattern inside the {} show how many times. The $ says that the max length is 9 and that there can't be more chars.
So you can read a condition like that:
([which chars are allowed]{How many chars are allowed})
^------------------------\/---------------------------^
One condition
And in Java you use it like that:
String test= "G12345678";
boolean isValid = Pattern.matches("([G]{1})([0-9]{8})$", test);
As you see that matches method takes two parameters. The first parameter is a regex and the second parameter is the string to check. If the string matches the pattern, it returns true.

Create an ArrayList. Ask the user to input the ID, check if it is already there in the list, ignore, otherwise add that ID to the list.
EDIT: For ensuring that the rest 8 characters of the String ID are digits, you can use the regex "\\d+". \d is for digits and + is for one or more digits.
Scanner sc = new Scanner(System.in);
ArrayList<String> IDS = new ArrayList();
char more = 'y';
String ID;
String regex = "\\d+";
while (more == 'y') {
System.out.println("Pleaes enter you ID.");
ID = sc.next();
if (IDS.contains(ID)) {
System.out.println("This ID is already added.");
} else if (ID.length() == 9 && ID.charAt(0) == 'G' && ID.substring(1).matches(regex)) {
IDS.add(ID);
System.out.println("Added");
} else {
System.out.println("Invalid ID");
}
System.out.println("Do you want to add more? y/n");
more = sc.next().charAt(0);
}

Assuming that you save the id as a string, you can check the first letter and then check if the rest is a number.
Ex.
String example = "G12345678";
String firstLetter = example.substring(0, 1); //this will give you the first letter
String number = example.substring(1, 9); //this will give you the number
To check that number is a number you could do the following instead of checking every character:
try {
foo = Integer.parseInt(number);
} catch (NumberFormatException e) {
//Will Throw exception!
//do something! anything to handle the exception.
// this is not a number
}

Breaking up a single user input and storing it in two different variables. (Java)

I'm very new to programming, especially Java. I need to create a program that counts how many orders each entry at a restaurant gets ordered. The restaurant carries 3 entries, hamburgers, salad, and special.
I need to set up my program so that the user inputs, say, "hamburger 3", it would keep track of the number and add it up at the end. If the user inputs "quit", the program would quit.
System.out.println("Enter the type (special, salad, or hamburger) of entrée followed by the number, or quit to exit the program.");
I'm thinking about using a while loop, setting it so if the user input != to "quit", then it would run.
What's difficult for me is I don't know how to make my program take into account the two different parts of the user input, "hamburger 3" and sum up the number part at the end.
At the end, I want it to say something like "You sold X hamburgers, Y salads, and Z specials today."
Help would be appreciated.

You'll probably want three int variables to use as a running tally of the number of orders been made:
public class Restaurant {
private int specials = 0;
private int salads = 0;
private int hamburger = 0;
You could then use a do-while loop to request information from the user...
String input = null;
do {
//...
} while ("quite".equalsIgnoreCase(input));
Now, you need some way to ask the user for input. You can use a java.util.Scanner easily enough for this. See the Scanning tutorial
Scanner scanner = new Scanner(System.in);
//...
do {
System.out.println("Enter the type (special, salad, or hamburger) of entrée followed by the number, or quit to exit the program.");
input = scanner.nextLine();
Now you have the input from the user, you need to make some decisions. You need to know if they entered valid input (an entree and an amount) as well as if they entered an available option...
// Break the input apart at the spaces...
String[] parts = input.split(" ");
// We only care if there are two parts...
if (parts.length == 2) {
// Process the parts...
} else if (parts.length == 0 || !"quite".equalsIgnoreCase(parts[0])) {
System.out.println("Your selection is invalid");
}
Okay, so we can now determine if the user input meets or first requirement or not ([text][space][text]), now we need to determine if the values are actually valid...
First, lets check the quantity...
if (parts.length == 2) {
// We user another Scanner, as this can determine if the String
// is an `int` value (or at least starts with one)
Scanner test = new Scanner(parts[1]);
if (test.hasInt()) {
int quantity = test.nextInt();
// continue processing...
} else {
System.out.println(parts[1] + " is not a valid quantity");
}
Now we want to check if the actually entered a valid entree...
if (test.hasInt()) {
int quantity = test.nextInt();
// We could use a case statement here, but for simplicity...
if ("special".equalsIgnoreCase(parts[0])) {
specials += quantity;
} else if ("salad".equalsIgnoreCase(parts[0])) {
salads += quantity;
} else if ("hamburger".equalsIgnoreCase(parts[0])) {
hamburger += quantity;
} else {
System.out.println(parts[0] + " is not a valid entree");
}
Take a look at The if-then and if-then-else Statements and The while and do-while Statements for more details.
You may also find Learning the Java Language of some help. Also, keep a copy of the JavaDocs at hand, it will make it eaiser to find references to the classes within the API

These two methods should be what you're looking for.
For splitting: String.split(String regex)
http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#split(java.lang.String)
For parsing String into an Interger: Integer.parseInt(String s)
http://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#parseInt(java.lang.String)

You can split your strings using input.split(" "). This method gives you two strings - two parts of the main string. The character you splitted with (" ") won't be found in the string anymore.
To then get an integer out of your string, you can use the static method Integer.parseInt(inputPartWithCount).
I hope this helps!