Develop Reference

Having an incedibly tough time with a loop

I've copied part of the instructions below, and I can code pretty much every part on its own, but getting the control flow together is giving me massive doubts about my ability.
One of my biggest problems is the int gameChanger. Im supposed to immediately verify if it is a integer or not, and loop back if its not. But then Im also supposed to check to see if thebuser ever types "exit". But the input variable for my scanner instance is an integer... So Im stumped. I can use a try catch to check the missmatchexception once the input is being read in, but that doesnt solve the exit issue nor am I able to come up with solid logic to get the try catch to loop back if it indeed isnt an integer. Im thinking a do while loop but I havent gotten it to work.
Instructions:

You can whether the input is a number before attempting to consume it.
int num;
while (true) {
if (scanner.hasNextInt()) {
num = scanner.nextInt();
break;
} else {
// read whatever is there instead.
String line = scanner.nextLine();
if (line.equals("exit"))
System.exit(0);
System.out.println("Please enter a number");
}
}
System.out.println("Number entered " + num);

This gets the job done. Try it out.
import java.util.Scanner;
public class MyCode
{
public static void main(String[] args)
{
String gameInput = ".";
int gameNumber = 0;
boolean inputLoop = true;
Scanner input = new Scanner(System.in);
while(inputLoop == true)
{
try
{
System.out.print("Please enter a valid game number: ");
gameInput = input.next();
if(gameInput.equals("exit"))
{
System.out.println("Program will now end. Goodbye.");
inputLoop = false;
input.close();
}
gameNumber = Integer.parseInt(gameInput);
if(gameNumber >= 20001 && gameNumber <= 21230)
{
System.out.println("You have inputted a valid game number.");
inputLoop = false;
input.close();
}
}
catch(NumberFormatException e)
{
if(!gameInput.equals("exit"))
{
System.err.println("Invalid game number. Please try again.");
}
}
}
}
}

About some simple exception handling in Java

There are several questions I would like to ask, please refer the comment part I have added in the code, Thanks.
package test;
import java.util.InputMismatchException;
import java.util.Scanner;
public class Test {
/* Task:
prompt user to read two integers and display the sum. prompt user to read the number again if the input is incorrect */
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean accept_a = false;
boolean accept_b = false;
int a;
int b;
while (accept_a == false) {
try {
System.out.print("Input A: ");
a = input.nextInt(); /* 1. Let's enter "abc" to trigger the exception handling part first*/
accept_a = true;
} catch (InputMismatchException ex) {
System.out.println("Input is Wrong");
input.nextLine(); /* 2. I am still not familiar with nextLine() parameter after reading the java manual, would you mind to explain more? All I want to do is "Clear Scanner Buffer" so it wont loop for the println and ask user to input A again, is it a correct way to do it? */
}
}
while (accept_b == false) {
try {
System.out.print("Input B: ");
b = input.nextInt();
accept_b = true;
} catch (InputMismatchException ex) { /*3. Since this is similar to the above situation, is it possible to reuse the try-catch block to handling b (or even more input like c d e...) exception? */
System.out.println("Input is Wrong");
input.nextLine();
}
}
System.out.println("The sum is " + (a + b)); /* 4. Why a & b is not found?*/
}
}

I am still not familiar with nextLine() parameter after reading the java manual, would you mind to explain more? All I want to do is "Clear Scanner Buffer" so it wont loop for the println and ask user to input A again, is it a correct way to do it?
The use of input.nextLine(); after input.nextInt(); is to clear the remaining content from the input stream, as (at least) the new line character is still in the buffer, leaving the contents in the buffer will cause input.nextInt(); to continue throwing an Exception if it's no cleared first
Since this is similar to the above situation, is it possible to reuse the try-catch block to handling b (or even more input like c d e...) exception?
You could, but what happens if input b is wrong? Do you ask the user to re-enter input a? What happens if you have 100 inputs and they get the last one wrong?You'd actually be better off writing a method which did this for, that is, one which prompted the user for a value and returned that value
For example...
public int promptForIntValue(String prompt) {
int value = -1;
boolean accepted = false;
do {
try {
System.out.print(prompt);
value = input.nextInt();
accepted = true;
} catch (InputMismatchException ex) {
System.out.println("Input is Wrong");
input.nextLine();
}
} while (!accepted);
return value;
}
Why a & b is not found?
Because they've not been initialised and the compiler can not be sure that they have a valid value...
Try changing it something more like.
int a = 0;
int b = 0;

Yes, it's okay. And will consume the non-integer input.
Yes. If we extract it to a method.
Because the compiler believes they might not be initialized.
Let's simplify and extract a method,
private static int readInt(String name, Scanner input) {
while (true) {
try {
System.out.printf("Input %s: ", name);
return input.nextInt();
} catch (InputMismatchException ex) {
System.out.printf("Input %s is Wrong%n", input.nextLine());
}
}
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int a = readInt("A", input);
int b = readInt("B", input);
System.out.println("The sum is " + (a + b));
}

I have put comment to that question line.
package test;
import java.util.InputMismatchException;
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean accept_a = false;
boolean accept_b = false;
int a=0;
int b=0;
System.out.print("Input A: ");
while (accept_a == false) {
try {
a = input.nextInt(); // it looks for integer token otherwise exception
accept_a = true;
} catch (InputMismatchException ex) {
System.out.println("Input is Wrong");
input.next(); // Move to next other wise exception // you can use hasNextInt()
}
}
System.out.print("Input B: ");
while (accept_b == false) {
try {
b = input.nextInt();
accept_b = true;
} catch (InputMismatchException ex) {
System.out.println("Input is Wrong");
input.next();
}
}
System.out.println("The sum is " + (a + b)); // complier doesn't know wheather they have initialised or not because of try-catch blocks. so explicitly initialised them.
}
}

Check out this "nextLine() after nextInt()"
and initialize the variable a and b to zero
nextInt() method does not read the last newline character.

Determining if number entered is an int [duplicate]

This question already has answers here:
What's the best way to check if a String represents an integer in Java?
(40 answers)
Closed 8 years ago.
import java.util.Scanner;
public class test {
/**
* #param args
*/
public static void main(String[] args)
{
Scanner input = new Scanner (System.in);
boolean US1 = false;
boolean game;
int score = 1;
int wage = 0;
int fin_score = 0;
String ans;
if (US1 == false) {
game = false;
System.out.println (score);
System.out.println("Enter a wager");
wage = input.nextInt();
}
if (wage < score) {
System.out.println ("What is the capital of Liberia?");
ans = input.next();
if (ans.equalsIgnoreCase("Monrovia")) {
System.out.println ("You got it right!");
System.out.println ("Final score " + fin_score);
}
}
}
}
I have found a bunch of solutions using InputMismatchException and try{}catch{} but they never work when they are implemented in my code. is there a way to implement these here? I am trying to make a loop that iterates until the wage entered is an integer

You can have multiple catch exceptions in your code to check for bad input. For example
try{
wage = input.nextInt();
catch (InputMismatchException e){
System.out.print(e.getMessage());
//handle mismatch input exception
}
catch (NumberFormatException e) {
System.out.print(e.getMessage());
//handle NFE
}
catch (Exception e) {
System.out.print(e.getMessage());
//last ditch case
}
Any of these would work fine for Scanner errors, but InputMismatchException is the best to use. It would help your case a great deal if you included the non-working code with the try-catch blocks.

First of all, You should be using Scanner.nextLine, because Scanner.nextInt uses spaces and newlines as delimiters, which is probably not what you want (any thing after a space will be left on the scanner, breaking any next reads).
Try this instead:
boolean valid = false;
System.out.print("Enter a wager: "); //Looks nicer when the input is put right next to the label
while(!valid)
try {
wage = Integer.valueOf(input.nextLine());
valid = true;
} catch (NumberFormatException e) {
System.out.print("That's not a valid number! Enter a wager: ");
}
}

Yes! There is a good way to do this:
Scanner input = new Scanner(System.in);
boolean gotAnInt = false;
while(!gotAnInt){
System.out.println("Enter int: ");
if(input.hasNextInt()){
int theInt = input.nextInt();
gotAnInt = true;
}else{
input.next();
}
}

Inputing Integers error throwing

Can someone help me make this code neater. I would rather use parse int than a buffer reader. I want my code to loop until the user inputs a number. I couldn't figure out how to do this without the code printing out the same statement twice.
public void setAge()
{
try {
age = Integer.parseInt(scan.nextLine());
} catch (NumberFormatException e) {
System.out.println("What is your age?");
this.setAge();
}
}
Alright, my question is unclear. I am unsure of how to handle the error that a scanner throws when you don't input an integer. How do I handle this? I found "NumberFormatException" in a different post, but I am unsure of what this does. Can anyone help me with this, or is my question still unclear?

Try this:
import java.util.InputMismatchException;
import java.util.Scanner;
public class TestScanner {
public static void main(String[] args) {
Scanner scanner = null;
int age = -1;
do {
try {
scanner = new Scanner(System.in);
System.out.println("What is your age?");
age = scanner.nextInt();
} catch (InputMismatchException e) {
System.out.println("Please enter a number!");
}
} while (age == -1);
System.out.println("You are " + age + " years old.");
if (scanner != null)
scanner.close();
}
}
I get this output (the first time I enter abc instead of a number to make it retry):
What is your age?
abc
Please enter a number!
What is your age?
35
You are 35 years old.
Have fun!

Use scan.nextInt(); instead of scan.nextLine();
With this, you don't need to parse the line.
EDIT: Oops, i misread your question

Number Format Exception occurs in the java code when a programmer tries to convert a String into a number. The Number might be int,float or any java numeric values.
The conversions are done by the functions Integer.parseInt.Consider if you give the value of str is "saurabh", the function call will fail to compile because "saurabh" is not a legal string representation of an int value and NumberFormatException will occurs

You could use a scanner.
You'll need to;
import java.util.*;
static Scanner console = new Scanner(System.in);
You won't need the parse statement at all.
age = console.nextInt();
EDIT: Editing my answer after seeing your edit.
I would put the entire try in a do loop. Using a new boolean variable to control when you come out of it.
boolean excep;
do {
excep = false;
try {
age = console.nextInt();
}
catch (Exception exRef) {
System.out.println("Please input an integer");
console.nextLine();
excep = true;
}
} while (excep);
The console.nextLine() just clears a line so it doesnt re-read the last input. Sometimes it's needed.
Using this i don't receive any error notifications on the running of it.

Try this:
static boolean firstTime = true;
public static void main(String[] args) {
boolean firstTime = true;
setAge();
}
public static void setAge()
{
if(firstTime)
{
System.out.println("What is your age?");
firstTime = false;
}
Scanner scan = new Scanner(System.in);
try{
int age = scan.nextInt();
System.out.println(age);
}
catch(InputMismatchException e)
{
setAge();
}
}

if you want to print different messages you would have to do like:
import java.util.Scanner;
public class Numbers {
public static void main(String args[]) {
Numbers numbers = new Numbers();
numbers.setAge();
}
private boolean alrearyAsked = false;
private int age = 0;
static Scanner scan = new Scanner(System.in);
public void setAge()
{
try {
age = scan.nextInt();
} catch (NumberFormatException e) {
if (alrearyAsked) {
System.out.println("you typed a wrong age, please try again.");
}
else {
System.out.println("What is your age?");
}
this.setAge();
}
}
}

What's the easiest way to check if inputed number is a positive integer and re-prompt if not?

The program takes user input which is supposed to be an integer greater than 0. If the user doesn't do this he is notified of the mistake and is reprompted. Once the correct input is entered, the value is returned. What's the best way to do this? The following code is my try but doesn't work. It seems unnecessarily complex for such an easy task.
System.out.println("Please enter an integer greater than 0:");
Scanner scan = new Scanner(System.in);
int red = -1;
do
{
try
{
red = scan.nextInt();
}catch(InputMismatchException e)
{
System.out.println("Number must be an integer");
scan.nextLine();
if(red < 1)
System.out.println("Number must be more than zero");
else
break;
}
}while(true);
return red;
Sometimes I don't know what to put in my question because I already know the code doesn't work - so if there's something else I should tell please let me know.

The basic concept is running in the right direction, beware though, nextInt won't consume the new line, leaving it within the scanner, meaning you will end up with an infinite loop after the first unsuccessful loop.
Personally, I would simply get the input as a String using nextLine, which will consume the new line, causing the next loop to stop at the statement.
Then I would simply parse the String to an int value using Integer.parseInt
For example...
Scanner scan = new Scanner(System.in);
int red = -1;
do {
System.out.print("Please enter an integer greater than 0:");
String text = scan.nextLine();
if (text != null && !text.isEmpty()) {
try {
red = Integer.parseInt(text);
// This is optional...
if (red < 1) {
System.out.println("Number must be more than zero");
}
} catch (NumberFormatException exp) {
// This is optional...
System.out.println("Not a number, try again...");
}
}
} while (red < 1);

I use this class instead of the Scanner or BufferedReader classes to get user input:
import java.io.*;
public class Input{
private static BufferedReader input=new BufferedReader
(new InputStreamReader(System.in));
public static Double getDouble(String prompt){
Double value;
System.out.print(prompt);
try{
value=Double.parseDouble(Input.input.readLine());
}
catch (Exception error){
// error condition
value=null;
}
return value;
}
public static Integer getInteger(String prompt){
Integer value;
System.out.print(prompt);
try{
value=Integer.parseInt(Input.input.readLine());
}
catch (Exception error){
// error condition
value=null;
}
return value;
}
public static String getString(String prompt){
String string;
System.out.print(prompt);
try{
string=Input.input.readLine();
}
catch (Exception error){
// error condition
string=null;
}
return string;
}
}
Now, to answer your question u can write your code like this:
public class InputTest {
public int checkValue() {
int value;
do {
value = Input.getInteger("Enter a value greater than 0: ");
} while (value <= 0);
return value;
}
}