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i want to display every number entered in my while loop.so how can i do this

must create a java application that will determine and display sum of numbers as entered by the user.The summation must take place so long the user wants to.when program ends the summation must be displayed as follows
e.g say the user enters 3 numbers
10 + 12+ 3=25
and you must use a while loop

Here's a function to do just that. Just call the function whenever you need.
Ex: System.out.println(parseSum("10 + 12+ 3")) → 25
public static int parseSum(String input) {
// Removes spaces
input = input.replace(" ", "");
int total = 0;
String num = "";
int letter = 0;
// Loop through each letter of input
while (letter < input.length()) {
// Checks if letter is a number
if (input.substring(letter, letter+1).matches(".*[0-9].*")) {
// Adds that character to String
num += input.charAt(letter);
} else {
// If the character is not a number, it turns the String to an integer and adds it to the total
total += Integer.valueOf(num);
num = "";
}
letter++;
}
total += Integer.valueOf(num);
return total;
}
The while loop is essentially a for loop though. Is there a specific reason why you needed it to be a while loop?

There is a lot of ways to achieve this. Here an example of code that could be improve (for example by catching an InputMismatchException if the user doesn't enter a number).
Please for the next time, post what you have tried and where you stuck on.
public static void main (String[] args) {
boolean playAgain = true;
while(playAgain) {
Scanner sc = new Scanner(System.in);
System.out.println("Please enter the first number : ");
int nb1 = sc.nextInt();
System.out.println("Ok! I got it! Please enter the second number : ");
int nb2 = sc.nextInt();
System.out.println("Great! Please enter the third and last number : ");
int nb3 = sc.nextInt();
int sum = nb1+nb2+nb3;
System.out.println("result==>"+nb1+"+"+nb2+"+"+nb3+"="+sum);
boolean validResponse = false;
while(!validResponse) {
System.out.println("Do you want to continue ? y/n");
String response = sc.next();
if(response.equals("n")) {
System.out.println("Thank you! see you next time :)");
playAgain = false;
validResponse = true;
} else if(response.equals("y")) {
playAgain = true;
validResponse = true;
} else {
System.out.println("Sorry, I didn't get it!");
}
}
}
}

How would i do input validation?

Tells the user if the number entered is even or even. I need help with the input validation. The validation i need do is that the user cannot entered anything but a number. Trying to do the validation without the try and catch method.
import java.util.Scanner;
public class oddoreven {
public static void main (String [] args) {
Scanner input = new Scanner (System.in);
//declaractions
int num;
//while loop
do{
System.out.println("PLease enter a number to see whether it is even or odd. To end tyype in -99.");
num = input.nextInt();
// input valid
}while(num != -99); // loop ends
// begins the method
public static void is_odd_or_even_number(int number){
int rem = number%2;
\

You can call Scanner.hasNextInt() to determine if the next input is an int (and consume anything else). Also, you might make an infinite loop and break when the input is -99 (or 99, your code tests for 99 but your prompt says -99). Finally, you should call your method. Something like,
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int num;
do {
System.out.println("Please enter a number to see whether it is "
+ "even or odd. To end type in -99.");
if (input.hasNextInt()) {
num = input.nextInt();
if (num != -99) { // <-- directions say -99.
is_odd_or_even_number(num);
} else {
break;
}
} else {
System.out.printf("%s is not a valid int.%n", input.nextLine());
}
} while (true);
}

You can use Scanner.nextLine() to get a string input. Then loop through the characters to make sure they are all digits. (assuming non-negative integers only)
string rawInput = input.nextLine();
boolean validInput = true;
for (char c : rawInput) {
if (!Character.isDigit(c)) {
validInput = false;
break;
}
}
if (validInput) {
int num == Integer.parseInt(rawInput);
// proceed as normal
}
else {
// invalid input, print out error message
}

You can use regex to check whether all the characters of string entered by user are digits or not,
num.matches("[0-9]+") // return true if all characters are digits
or
num.matches("^[0-9]*$") // return true if all characters are digits
but before that change your num = input.nextint() to num = nextLine() and make num as String. if you dont do this there is no need of validating user input as you are requiring.

Using java scanner to check two conditions while taking user input

I need to user to enter an int between 1 and 301.
I have this simple loop here to check for user input.
I just want a single number from the user, and if the user enters anything other than an int between 1 and 301, I want to display the print line and prompt the users to try again until they enter a valid input.
while (!sc.hasNextInt()) {
System.out.print("Invalid Input. Please enter a valid number between 1 and 301: ");
sc.next();
}
int numToCheck = sc.nextInt();
// do stuff with numToCheck
This checks that the input is an int, but I can't seem to find a way to give the int input a bound. I tried to assign the user input to a variable and then check the conditions input < 1 or input > 301, but I get InputMismatchException if user enters a letter. How should I store the user input? (I want to store it as an int to check the conditions, but can't do that since I don't know what the user will enter).
Perhaps there is a better design to accomplish all this. Those are welcomed too.
Thanks in advance.

You're not saving the value of the of the input. So your program is waiting on the user to enter a number each time it see "sc.nextInt()" Assign the input to a variable, and then check the condition.
EDIT: okay, I'll go the extra mile for you. See if this works.
***Accounted for the case where the user might enter a character instead of a number.
import java.util.*;
public class HelloWorld{
public static void main(String []args){
Scanner sc = new Scanner(System.in);
int input;
while (true){
if (sc.hasNextInt()){
input = sc.nextInt(); // Assign the next integer to a variable
if (input <= 301 && input >= 1){ // Check if integer meets condition
break; // Condition met, break out of loop
}
}else{
sc.next();
}
System.out.println("Invalid Input. Please enter a valid number between 1 and 301: ");
}
}
}

I ran this code, to see if it would show a better performance than yours.
Scanner sc = new Scanner(System.in);
boolean valid = true;
do {
if (!valid) {
System.out.print("Invalid Input. ");
}
System.out.print("Please enter a valid number between 1 and 301: ");
String input = sc.next();
try {
int value = Integer.parseInt(input);
valid = (value >= 1 && value <= 301);
} catch (NumberFormatException nfex) {
valid = false;
}
} while (!valid);
When the conversion to integer fails, the JVM hangs a little. I believe your problem has more to do with the try / catch mecanism that Scanner performs under the hood, than with design.

Assuming you want only 1 input from the user, try following simple code, which takes input from the user until user enters a valid input.
Scanner in = new Scanner(System.in);
int flag = 0,x=0;
while(flag == 0){
x = in.nextInt();
if(x<1 || x>301){
flag = 0;
System.out.println("Invalid Input.");
}
else{
flag = 1;
}
}
And if you want user to input more than 1 inputs (i.e 3 here), than set a counter that increases with every valid input of the user, as following:
Scanner in = new Scanner(System.in);
int flag = 0,x=0,count = 1;
while(flag == 0){
x = in.nextInt();
if(x<1 || x>301){
flag = 0;
System.out.println("Invalid Input.");
}
else{
//executes when input is valid
if(count == 3){
flag = 1;
}
count++;
}
}
Edit:
If you also want to check whether the input is Integer or not, than you have to add one extra condition in above code. And as you said you want only one input from user rather than 3, you have to change exit condition. Change code as following:
Scanner in = new Scanner(System.in);
int flag = 0,count = 1,x=0,flag1 = 0;
String y;
while(flag == 0){
y = in.next();
flag1 = 0;
try{
x = Integer.parseInt(y);
}
catch(NumberFormatException e){
flag1 = 1;
System.out.println("Invalid Input.");
}
if((x<1 || x>301)&&flag1 == 0){
flag = 0;
System.out.println("Invalid Input.");
}
else if(flag1 == 0){
//executes when input is valid
if(count == 1){ // put count == 3 if you want 3 inputs from user.
flag = 1;
}
count++;
}
}
Here we are taking the input as a String and than converting the String into the Integer by using Integer.parseInt(). If the String is not Integer, than it will throw the exception and we will continue the loop till the valid input is entered by the user.

Use DO WHILE for result
do{
System.out.print("value of x : " + x );
x++;
System.out.print("\n");
}while( x < 20 );
OK ?

A tiny issue with exception handling interactment - not the correct output

package test5555;
import java.util.InputMismatchException;
import java.util.Random;
import java.util.Scanner;
public class Test5555 {
private static int[] randomInteger;
public static void main(String[] args) {
boolean validInput = false;
randomInteger = new int[100];
Random rand = new Random();
for (int i = 0; i < randomInteger.length; i++)
randomInteger[i] = rand.nextInt();
int indexPosition = 0;
Scanner input = new Scanner(System.in); {
System.out.println("Please enter an integer for the array index position: ");
while(!validInput)
{
try
{
indexPosition = input.nextInt();
validInput = true;
System.out.println(randomInteger[indexPosition]);
} catch ( InputMismatchException | IndexOutOfBoundsException ex) {
System.out.print("Please enter a valid integer between 0 and 100 or type quit to exit: ");
String s = input.next();
if(s.equals("quit")){
System.exit(0);
System.out.println(randomInteger[indexPosition]);
}
}
}
}
}
}
The code runs perfectly except for two minor hiccups that I cannot solve. When you run it you get Please enter an integer for the array index position:If you type a number above 100 or a string such as bob then you get Please enter a valid integer between 0 and 100 or type quit to exit:which is perfect. But if you type quit then you get Please enter a valid integer between 0 and 100 or type quit to exit: BUILD SUCCESSFUL (total time: 2 minutes 2 seconds) so it quits it but it repeats the exception statement which I do not want.
When you type a number above 100 and receive the Please enter a valid integer between 0 and 100 or type quit to exit: if you then type a correct integer the program will just turn off and it will say BUILD SUCCESSFUL instead of retrieving the number for you from the array

Replace your while loop code with below,
String s = null;
while(!validInput)
{
try
{
if(s != null){
indexPosition = Integer.parseInt(s);
}
else{
indexPosition = input.nextInt();
}
System.out.println(randomInteger[indexPosition]);
validInput = true;
} catch ( InputMismatchException | NumberFormatException | IndexOutOfBoundsException ex ) {
System.out.println("Please enter a valid integer between 0 and 100 or type quit to exit: ");
input.nextLine();
s = input.next();
if(s.equals("quit")){
System.exit(0);
}
}
}
Please read this to get more idea on Scanner.
https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html
In your case the problem is (As per doc)
When a scanner throws an InputMismatchException, the scanner will not
pass the token that caused the exception, so that it may be retrieved
or skipped via some other method.

The behavior you describe in point 1 us not correct. If you type in a number and then quit it works "as expected"
If you type in a string such as "bob" your nextInt() fails with an InputMissmatchException which means your "input.next()" call in the catch clause will read "bob" and see it's not equal to "quit" and just go back to the loop and block and wait for an "int".
In point 2. You type an int and you get an exception...but you've set validInput to true already so you'll exit the loop. You need to set validInput after you print.

If I got your question correctly, I would implement little differently. Please check if it fulfills your requirement.
import java.util.Random;
import java.util.Scanner;
public class Test5555 {
private static int[] randomInteger;
public static void main(String[] args) {
randomInteger = new int[100];
Random rand = new Random();
int indexPosition;
for (int i = 0; i < randomInteger.length; i++)
randomInteger[i] = rand.nextInt();
Scanner input = new Scanner(System.in);
System.out.println("Please enter an integer for the array index position: ");
while(true) {
String strIndex = input.next();
if(strIndex.equals("quit")) break;
indexPosition = getIntVal(strIndex);
if(indexPosition < 0 || indexPosition >= randomInteger.length) {
System.out.print("Please enter a valid integer between 0 and "
+ randomInteger.length + " or type quit to exit: ");
continue;
}
System.out.println(randomInteger[indexPosition]);
break;
}
input.close();
}
protected static int getIntVal(String inputStr) {
int result = -1;
try {
result = Integer.parseInt(inputStr);
} catch(NumberFormatException e) {}
return result;
}
}

This part of your code is wrong
try
{
indexPosition = input.nextInt();
validInput = true;
System.out.println(randomInteger[indexPosition]);
} catch ( InputMismatchException | IndexOutOfBoundsException ex) {
You are saying that your indexPosition is right before to check it, the line
validInput = true;
Should be later of check if the array have that position.
Right code:
...
indexPosition = input.nextInt();
System.out.println(randomInteger[indexPosition]);
validInput = true;
....

Try/Catch inside While in static main method

I don't understand the logic to this. If I run this code and enter a non-int such as the letter f, I get stuck in an infinite loop outputting the two println's and I am not given another chance to input an int to the scanner...it just keeps spitting out words to the console.
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);//<<<<<SCANNER HERE
int opponents = 0;
boolean opponentsCreated = false;
while(opponentsCreated == false)
{
try
{
System.out.print("How many players: ");
int tempOpponents = scan.nextInt();
if(tempOpponents > 0)
{
opponents = tempOpponents;
opponentsCreated = true;
}
}
catch(InputMismatchException notAValidInt)
{
System.out.println("Not valid - must be a number greater than 0 ");
}
}
}
But if I simply change the Scanner to be declared inside the while loop, all of a sudden the program works as expected:
public static void main(String[] args) {
int opponents = 0;
boolean opponentsCreated = false;
while(opponentsCreated == false)
{
Scanner scan = new Scanner(System.in);//<<<<<SCANNER HERE
try
{
System.out.print("How many players: ");
int tempOpponents = scan.nextInt();
if(tempOpponents > 0)
{
opponents = tempOpponents;
opponentsCreated = true;
}
}
catch(InputMismatchException notAValidInt)
{
System.out.println("Not valid - must be a number greater than 0 ");
}
}
}
I honestly just sat here for 2 hours trying to figure out what the heck was wrong with my program only to find out it was a matter of where I declared my Scanner even though in both versions of the code the Scanner was not out of scope. So now I'm really curious why it works this way

Adding on to #HovercraftFullOfEels answer:
The root cause is, the scanner position does not move in case of the said exception. So scanner keeps reating same bad input again and again. Quoting JavaDoc
If the translation is successful, the scanner advances past the input
that matched.
catch(InputMismatchException notAValidInt)
{
scan.reset();
System.out.println("Not valid - must be a number greater than 0 ");
//position is still 0
scan.next(); //position is now 1
}
To visualize:
Input: f______________
Scanner position: ^______________
InputMismatchException ^______________
scan.next() _^_____________
Relevant source (look at the source comment):
try {
String s = next(integerPattern());
if (matcher.group(SIMPLE_GROUP_INDEX) == null)
s = processIntegerToken(s);
return Integer.parseInt(s, radix);
} catch (NumberFormatException nfe) {
position = matcher.start(); // don't skip bad token
throw new InputMismatchException(nfe.getMessage());
}

One possible problem is that you may be leaving the end of line token hanging when an excpetion occurs. If you handle this by making sure to swallow the end of line token when needed, you are likely OK. For example:
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);// <<<<<SCANNER HERE
int opponents = 0;
boolean opponentsCreated = false;
while (opponentsCreated == false) {
try {
System.out.print("How many players: ");
int tempOpponents = scan.nextInt();
// line below corrected!
scan.nextLine(); // *** this might not be a bad idea either ***
if (tempOpponents > 0) {
opponents = tempOpponents;
opponentsCreated = true;
}
} catch (InputMismatchException notAValidInt) {
System.out.println("Not valid - must be a number greater than 0 ");
scan.nextLine(); // ****** this is what you need here *****
}
}
}
Nice question, by the way!