I'm trying to make a small program more robust and I need some help with that.
Scanner kb = new Scanner(System.in);
int num1;
int num2 = 0;
System.out.print("Enter number 1: ");
num1 = kb.nextInt();
while(num2 < num1) {
System.out.print("Enter number 2: ");
num2 = kb.nextInt();
}
Number 2 has to be greater than number 1
Also I want the program to automatically check and ignore if the user enters a character instead of a number. Because right now when a user enters for example r instead of a number the program just exits.
Use Scanner.hasNextInt():
Returns true if the next token in this scanner's input can be interpreted as an int value in the default radix using the nextInt() method. The scanner does not advance past any input.
Here's a snippet to illustrate:
Scanner sc = new Scanner(System.in);
System.out.print("Enter number 1: ");
while (!sc.hasNextInt()) sc.next();
int num1 = sc.nextInt();
int num2;
System.out.print("Enter number 2: ");
do {
while (!sc.hasNextInt()) sc.next();
num2 = sc.nextInt();
} while (num2 < num1);
System.out.println(num1 + " " + num2);
You don't have to parseInt or worry about NumberFormatException. Note that since the hasNextXXX methods don't advance past any input, you may have to call next() if you want to skip past the "garbage", as shown above.
Related questions
How do I keep a scanner from throwing exceptions when the wrong type is entered? (java)
the condition num2 < num1 should be num2 <= num1 if num2 has to be greater than num1
not knowing what the kb object is, I'd read a String and then trying Integer.parseInt() and if you don't catch an exception then it's a number, if you do, read a new one, maybe by setting num2 to Integer.MIN_VALUE and using the same type of logic in your example.
This should work:
import java.util.Scanner;
public class Test {
public static void main(String... args) throws Throwable {
Scanner kb = new Scanner(System.in);
int num1;
System.out.print("Enter number 1: ");
while (true)
try {
num1 = Integer.parseInt(kb.nextLine());
break;
} catch (NumberFormatException nfe) {
System.out.print("Try again: ");
}
int num2;
do {
System.out.print("Enter number 2: ");
while (true)
try {
num2 = Integer.parseInt(kb.nextLine());
break;
} catch (NumberFormatException nfe) {
System.out.print("Try again: ");
}
} while (num2 < num1);
}
}
Try this:
public static void main(String[] args)
{
Pattern p = Pattern.compile("^\\d+$");
Scanner kb = new Scanner(System.in);
int num1;
int num2 = 0;
String temp;
Matcher numberMatcher;
System.out.print("Enter number 1: ");
try
{
num1 = kb.nextInt();
}
catch (java.util.InputMismatchException e)
{
System.out.println("Invalid Input");
//
return;
}
while(num2<num1)
{
System.out.print("Enter number 2: ");
temp = kb.next();
numberMatcher = p.matcher(temp);
if (numberMatcher.matches())
{
num2 = Integer.parseInt(temp);
}
else
{
System.out.println("Invalid Number");
}
}
}
You could try to parse the string into an int as well, but usually people try to avoid throwing exceptions.
What I have done is that I have defined a regular expression that defines a number, \d means a numeric digit. The + sign means that there has to be one or more numeric digits. The extra \ in front of the \d is because in java, the \ is a special character, so it has to be escaped.
I see that Character.isDigit perfectly suits the need, since the input will be just one symbol.
Of course we don't have any info about this kb object but just in case it's a java.util.Scanner instance, I'd also suggest using java.io.InputStreamReader for command line input. Here's an example:
java.io.BufferedReader reader = new java.io.BufferedReader(new java.io.InputStreamReader(System.in));
try {
reader.read();
}
catch(Exception e) {
e.printStackTrace();
}
reader.close();
What you could do is also to take the next token as a String, converts this string to a char array and test that each character in the array is a digit.
I think that's correct, if you don't want to deal with the exceptions.
Related
So far I have this:
public double checkValueWithin(int min, int max) {
double num;
Scanner reader = new Scanner(System.in);
num = reader.nextDouble();
while (num < min || num > max) {
System.out.print("Invalid. Re-enter number: ");
num = reader.nextDouble();
}
return num;
}
and this:
public void askForMarks() {
double marks[] = new double[student];
int index = 0;
Scanner reader = new Scanner(System.in);
while (index < student) {
System.out.print("Please enter a mark (0..30): ");
marks[index] = (double) checkValueWithin(0, 30);
index++;
}
}
When I test this, it can't take double number and I got this message:
Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Scanner.java:909)
at java.util.Scanner.next(Scanner.java:1530)
at java.util.Scanner.nextDouble(Scanner.java:2456)
at MarkingSystem.checkValueWithin(MarkingSystem.java:25)
at MarkingSystem.askForMarks(MarkingSystem.java:44)
at World.main(World.java:6)
Java Result: 1
How do I fix this?
Instead of using a dot, like: 1.2, try to input like this: 1,2.
Here you can see the nature of Scanner:
double nextDouble()
Returns the next token as a double. If the next token is not a float or
is out of range, InputMismatchException is thrown.
Try to catch the exception
try {
// ...
} catch (InputMismatchException e) {
System.out.print(e.getMessage()); //try to find out specific reason.
}
UPDATE
CASE 1
I tried your code and there is nothing wrong with it. Your are getting that error because you must have entered String value. When I entered a numeric value, it runs without any errors. But once I entered String it throw the same Exception which you have mentioned in your question.
CASE 2
You have entered something, which is out of range as I have mentioned above.
I'm really wondering what you could have tried to enter. In my system, it is running perfectly without changing a single line of code. Just copy as it is and try to compile and run it.
import java.util.*;
public class Test {
public static void main(String... args) {
new Test().askForMarks(5);
}
public void askForMarks(int student) {
double marks[] = new double[student];
int index = 0;
Scanner reader = new Scanner(System.in);
while (index < student) {
System.out.print("Please enter a mark (0..30): ");
marks[index] = (double) checkValueWithin(0, 30);
index++;
}
}
public double checkValueWithin(int min, int max) {
double num;
Scanner reader = new Scanner(System.in);
num = reader.nextDouble();
while (num < min || num > max) {
System.out.print("Invalid. Re-enter number: ");
num = reader.nextDouble();
}
return num;
}
}
As you said, you have tried to enter 1.0, 2.8 and etc. Please try with this code.
Note : Please enter number one by one, on separate lines. I mean, enter 2.7, press enter and then enter second number (e.g. 6.7).
I encountered the same problem.
Strange, but the reason was that the object Scanner interprets fractions depending on localization of system.
If the current localization uses a comma to separate parts of the fractions, the fraction with the dot will turn into type String.
Hence the error ...
Since you have the manual user input loop, after the scanner has read your first input it will pass the carriage/return into the next line which will also be read; of course, that is not what you wanted.
You can try this
try {
// ...
} catch (InputMismatchException e) {
reader.next();
}
or alternatively, you can consume that carriage return before reading your next double input by calling
reader.next()
Are you providing write input to the console ?
Scanner reader = new Scanner(System.in);
num = reader.nextDouble();
This is return double if you just enter number like 456.
In case you enter a string or character instead,it will throw java.util.InputMismatchException when it tries to do num = reader.nextDouble() .
How to have a try and catch statement to validate the input in variable odd is a number else print an error message? I am new in java programming. Appreciate any help
import java.util.Scanner;
public class ScenarioB{
public static void main (String[]args){
int odd;
Scanner scan = new Scanner(System.in);
System.out.print("Enter An Odd Number\n");
odd = scan.nextInt();
}
}
The Scanner class has a method for this
Scanner in = new Scanner(System.in);
int num;
if(in.hasNextInt()){
num = in.nextInt();
System.out.println("Valid number");
}else{
System.out.println("Not a valid number");
}
Output 1:
8
Valid number
Output 2:
a
Not a valid number
Your code will like bellow:
try {
int scannedNumber;
Scanner scan = new Scanner(System.in);
System.out.print("Enter An Odd Number\n");
String userInput = scan.nextLine();
System.out.println("You pressed :" + userInput);
scannedNumber = Integer.parseInt(userInput);
if (scannedNumber % 2 == 0) {
throw new Exception("Not an odd number");
}
System.out.println("You pressed odd number");
} catch (Exception e) {
System.out.println("" + e.getMessage());
}
Here after taking input as String. Then cast as integer. If casting failed then it will throw NumberFormatException. If it is a number then check is it odd? If not odd then it is also throwing an exception.
Hope this will help you.
Thanks :)
For a college assessment I'm having to use a Scanner called sc with a class-level scope, and the entirety of the program has to be contained in a single class. The main method calls a menu() method, which uses the Scanner and a for loop to call one of two methods in response to user input.
One of the two methods uses the Scanner to calculate the factorial of an input integer. Once the method is executed, the for loop in menu() continues. To avoid an InputMismatchException due to the user entering a float, I used try/catch. However when the program returns back to the menu() for loop the Scanner causes an InputMismatchException when assigning to choice. How can I get Scanner to prompt the user for input again? Apologies if I'm missing something obvious, this is the first programming language I've ever learned. This should be the stripped down compilable code:
package summativeassessment;
import java.util.InputMismatchException;
import java.util.Scanner;
public class SummativeAssessment {
private static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
menu();
}
public static void menu(){
String fName;
String sName;
System.out.print("Enter your first name: ");
fName = sc.next();
System.out.print("Enter your last name: ");
sName = sc.next();
try{
for(int choice = 1; choice!=0;){
System.out.print("Option 1 to generate username. Option 2 to calculate factorial. Press 0 to quit: ");
choice = sc.nextInt();
switch(choice){
case 2:
System.out.println(fName+" "+sName+", you have selected option 2");
numberFactorial();
break;
case 0:
break;
default:
System.out.println("Invalid option. Please try again.");
}
}
} catch(InputMismatchException ex){
String msg = ex.getMessage();
System.out.println(msg);
}
}
public static void numberFactorial(){
System.out.print("Enter a number: ");
try{
int numIn = sc.nextInt();
long result = numIn;
if(numIn>0){
for(int factor = 1; factor<numIn; factor++){
result *= factor;
if(factor==numIn-1){
System.out.println("The factorial is "+result);
}
}
}
else{
System.out.println("Enter a positive integer greater than 0");
}
}
catch(InputMismatchException ex){
System.out.println("Input invalid");
}
}
}
I debugged your code and got this result:
If you enter a float as input you trigger the InputMismatchException but there is still something in your buffer. So the next time sc.nextInt() is called, it won't wait until you input a value because something is in the buffer already, so it takes the next value out of the buffer and tries to interpret is as an integer. However, it fails to do so, because it is not an integer, so an InputMismatchException is raised again and caught in your menu's catch, now leading to the exit of the program.
The solution is to draw whatever is left in the buffer after the exception was raised the first time.
So the working code will contain a buffer clearing sc.next() inside the exception:
public static void numberFactorial(){
System.out.print("Enter a number: ");
try{
int numIn = sc.nextInt();
long result = numIn;
if(numIn>0){
for(int factor = 1; factor<numIn; factor++){
result *= factor;
if(factor==numIn-1){
System.out.println("The factorial is "+result);
}
}
}
else{
System.out.println("Enter a positive integer greater than 0");
}
}
catch(InputMismatchException ex){
System.out.println("Input invalid");
sc.next();
}
}
So I'm trying to make a user input calculator using try-catch and exception. The program keeps on repeating and couldn't get the actual input for the numbers and it also includes the statement Wrong input. Please try again.
Any idea on how to fix this?
import java.util.*;
public class Calculator
{
public static void main(String[] args) {
int x=1;
do {
try {
System.out.println("Menu");
System.out.println("1-Addition");
System.out.println("2-Subtraction");
System.out.println("3-Multiplication");
System.out.println("4-Divison");
System.out.println("5-Modulos");
System.out.println("6-Exit");
System.out.println("Choose option: ");
Scanner scan = new Scanner(System.in);
int choice = scan.nextInt();
switch (choice) {
case 1:
System.out.print("Input two numbers:");
String dimension = scan.nextLine();
String[] parts = dimension.split(" ");
int a = Integer.parseInt(parts[0]);
int b = Integer.parseInt(parts[1]);
int c=a+b;
System.out.println("Sum = " +c);
break;
case 2:
System.out.print("Input two numbers:");
String dif = scan.nextLine();
String[] difference = dif.split(" ");
int num1 = Integer.parseInt(difference[0]);
int num2 = Integer.parseInt(difference[1]);
int d=num1-num2;
System.out.println("Difference = " +d);
break;
case 3:
System.out.print("Input two numbers:");
String multi = scan.nextLine();
String[] product = multi.split(" ");
int num3 = Integer.parseInt(product[0]);
int num4 = Integer.parseInt(product[1]);
int p=num3*num4;
System.out.println("Product = " +p);
break;
case 4:
System.out.print("Input two numbers:");
String div = scan.nextLine();
String[] quotient = div.split(" ");
int num5 = Integer.parseInt(quotient[0]);
int num6 = Integer.parseInt(quotient[1]);
int q=num5/num6;
System.out.println("Quotient = " +q);
break;
case 5:
System.out.print("Input two numbers:");
String mod = scan.nextLine();
String[] modulo = mod.split(" ");
int num7 = Integer.parseInt(modulo[0]);
int num8 = Integer.parseInt(modulo[1]);
int m=num7%num8;
System.out.println("Modulos = " +m);
break;
case 6:
System.out.println("Now exiting program...");
break;
}
} catch (Exception e) {
System.out.println("Wrong input. Try again.");
}
} while (x==1);
}
}
You need to do this
int choice = Integer.parseInt(scan.nextLine());
because you are reading next input with readline()
String dimension = scan.nextLine();
and \n is already present it the stream when you enter your option with nextInt() because nextint() never reads the \n which you leave behind by pressing Enter button.
You should add:
scan.skip("\n");
right after
int choice = scan.nextInt();
Why? Because nextInt() reads next integer from input. And it leaves new line character at the end on the input. It's like:
1<enter>
After that you invoke scan.nextLine() which reads everything up to next new line character. In fact there is 1 new line character in the buffer. And here your nextLine() reads an empty string.
To solve that issue you have to skip buffered new line characters with scan.skip("\n").
EDIT:
in addition, your code doesn't allow you to exit, because you're not changing your x variable. Try to change it after System.out.println("Now exiting program...");. It will work ;)
This will help you to identify root cause:
} catch (Exception e) {
System.out.println("Wrong input. Try again.");
e.printStackTrace();
}
I'm new to Java and I wanted to keep on asking for user input until the user enters an integer, so that there's no InputMismatchException. I've tried this code, but I still get the exception when I enter a non-integer value.
int getInt(String prompt){
System.out.print(prompt);
Scanner sc = new Scanner(System.in);
while(!sc.hasNextInt()){
System.out.println("Enter a whole number.");
sc.nextInt();
}
return sc.nextInt();
}
Thanks for your time!
Take the input using next instead of nextInt. Put a try catch to parse the input using parseInt method. If parsing is successful break the while loop, otherwise continue.
Try this:
System.out.print("input");
Scanner sc = new Scanner(System.in);
while (true) {
System.out.println("Enter a whole number.");
String input = sc.next();
int intInputValue = 0;
try {
intInputValue = Integer.parseInt(input);
System.out.println("Correct input, exit");
break;
} catch (NumberFormatException ne) {
System.out.println("Input is not a number, continue");
}
}
Shorter solution. Just take input in sc.next()
public int getInt(String prompt) {
Scanner sc = new Scanner(System.in);
System.out.print(prompt);
while (!sc.hasNextInt()) {
System.out.println("Enter a whole number");
sc.next();
}
return sc.nextInt();
}
Working on Juned's code, I was able to make it shorter.
int getInt(String prompt) {
System.out.print(prompt);
while(true){
try {
return Integer.parseInt(new Scanner(System.in).next());
} catch(NumberFormatException ne) {
System.out.print("That's not a whole number.\n"+prompt);
}
}
}
Keep gently scanning while you still have input, and check if it's indeed integer, as you need:
String s = "This is not yet number 10";
// create a new scanner
// with the specified String Object
Scanner scanner = new Scanner(s);
while (scanner.hasNext()) {
// if the next is a Int,
// print found and the Int
if (scanner.hasNextInt()) {
System.out.println("Found Int value :"
+ scanner.nextInt());
}
// if no Int is found,
// print "Not Found:" and the token
else {
System.out.println("Not found Int value :"
+ scanner.next());
}
}
scanner.close();
As an alternative, if it is just a single digit integer [0-9], then you can check its ASCII code. It should be between 48-57 to be an integer.
Building up on Juned's code, you can replace try block with an if condition:
System.out.print("input");
Scanner sc = new Scanner(System.in);
while (true) {
System.out.println("Enter a whole number.");
String input = sc.next();
int intInputValue = 0;
if(input.charAt(0) >= 48 && input.charAt(0) <= 57){
System.out.println("Correct input, exit");
break;
}
System.out.println("Input is not a number, continue");
}