This question already has answers here:
How to handle infinite loop caused by invalid input (InputMismatchException) using Scanner
(5 answers)
Closed 4 years ago.
Im trying to create a simple program where user enters an integer and if its not an integer it prints that an error occured and the program loops until user enters an integer. I found this try-catch solution but it doesn`t work properly. If user doesnt enter an integer the program will infinitly loop.
What would the correct solution be?
Scanner input = new Scanner(System.in);
int number;
boolean isInt = false;
while(isInt == false)
{
System.out.println("Enter a number:");
try
{
number = input.nextInt();
isInt = true;
}
catch (InputMismatchException error)
{
System.out.println("Error! You need to enter an integer!");
}
}
You're close.
It's easier to fail on parsing a string than it is to try to get an int from the Scanner since the scanner will block until it gets an int.
Scanner input = new Scanner(System.in);
int number;
boolean isInt = false;
while (isInt == false) {
System.out.println("Enter a number:");
try {
number = Integer.parseInt(input.nextLine());
isInt = true;
} catch (NumberFormatException error) {
System.out.println("Error! You need to enter an integer!");
}
}
Related
This question already has answers here:
Validating input using java.util.Scanner [duplicate]
(6 answers)
Closed 6 years ago.
I want to say enter an integer when someone trying to enter a string in this code.
Can you help me?
Here is my code:
import java.util.Scanner;
public class kl {
public static void main(String[] args) {
boolean primen = true;
Scanner input = new Scanner(System.in);
System.out.print("Please enter a positive integer that is prime or not : ");
int ncheck = input.nextInt();
if (ncheck < 2) {
primen = false;
}
for (int i = 2; i < ncheck; i++) {
if (ncheck % i == 0) {
primen = false;
break;
}
}
if (primen == true) {
System.out.println(ncheck + " is a prime number.");
}
else {
System.out.println(ncheck + " is not a prime number.");
}
}
}
You can find your solution here: Exception handling, Or use codes below
Here is your complete code:
while(true){
System.out.print("Please enter a positive integer that is prime or not : ");
try{
int i = input.nextInt();
break;
}catch(InputMismatchException e){
System.out.print("Wrong type input, pls try again!");
input.nextLine(); \\ prevent infinite loop
}
You can see: I use a Exception handle processor to catch the InputMismatchException and print on console the message. You can replace InputMismatchException by Exception. It's largest Exception Handler class in java
There are two approaches you can use with Scanner.
Call nextInt() and then catch and handle the InputMismatchException that you will get if the next input token isn't an integer.
Call hasNextInt(). If that returns true then call nextInt().
In either case, if you expected an integer and the user entered something else, then neither nextInt() or hasNextInt() will "consume" the unexpected characters. So if you want the user to try try again, you need to call nextLine() which will read all remaining characters on the line. You will typically discard them.
For more information on handling exceptions:
https://docs.oracle.com/javase/tutorial/essential/exceptions/handling.html
For more information on using Scanner:
https://docs.oracle.com/javase/tutorial/essential/io/scanning.html
http://docs.oracle.com/javase/8/docs/api/java/util/Scanner.html
This question already has answers here:
Error catching with try-catch and while loop [duplicate]
(1 answer)
How to handle infinite loop caused by invalid input (InputMismatchException) using Scanner
(5 answers)
Closed 7 years ago.
Hi I´m newbie in programming and I have this problem. I want to get input from user , using scanner. Program is supposed to try if input is valid, then call a function to do the job. Problem is I want program to repeat aking for input from user, if input is not valid. So I have try block in a while loop. The problem is on the first iteration of while loop everything is ok but when I insert invalid input and while loop is forced to iterate second time, try block is not executed and boolean which is condition of while loop is not set to false. So while loop runs for ever. Plz be kind.
public static void main(String args[]) {
Scanner scanner = new Scanner(System.in);
boolean isRunning = true;
int input;
while(isRunning) {
System.out.println("insert a number:");
try {
input = scanner.nextInt();
inputToString(input);
isRunning = false;
} catch(InputMismatchException e) {
System.out.println("input musi byt cele cislo");
isRunning = true;
}
}
}
public static void inputToString(int input) {
System.out.println(input);
}
You need to consume invalid number from Scanner before asking for new one since nextInt didn't consume it, but thrown exception.
} catch (InputMismatchException e) {
System.out.println("input musi byt cele cislo");
scanner.next();//consume
isRunning = true;
}
But you should not use exceptions and try-catch sections as main part of control logic. Scanner provides hasNextInt() method to test if user provided valid input. So your code should be like
System.out.print("give me the money:");//ask user of integer
while(!scanner.hasNextInt()){//test is provided value is valid integer
String token = scanner.next();//consume incorrect value
//inform that value is not correct and ask for new one
System.our.println(token + " is not considered as valid number");
System.our.print("please try again:");
}
//here we know that user provided valid integer
value = scanner.nextInt();
The problem with your approach is that when the input is not a valid int you leave it in the input buffer. The next loop will see exactly the same input, and it would repeat exactly the same actions (throw, catch, continue with the loop).
What you should do is to drop the incorrect input from the scanner when you see that it does not match what you expect. You could do it in your catch block, like this:
try {
input = scanner.nextInt();
inputToString(input);
isRunning = false;
} catch(InputMismatchException e) {
System.out.println("input musi byt cele cislo");
isRunning = true;
scanner.nextLine(); // Drop input from the scanner's buffer
}
A better approach would be to not rely on try/catch in the scanning code at all. Scanner provides a convenient way for you to avoid exceptions: calling hasNextInt before you call nextInt lets you find out ahead of time if the exception would be thrown or not, and clean the wrong data from the buffer:
while(isRunning) {
System.out.println("insert a number:");
if (scanner.hasNextInt()) {
input = scanner.nextInt();
inputToString(input);
isRunning = false;
} else {
System.out.println("input musi byt cele cislo");
isRunning = true;
scanner.nextLine(); // Drop input from the scanner's buffer
}
}
you should reinitialize Scanner reference before scanning input
public static void main(String args[]) {
Scanner scanner = null;
boolean isRunning = true;
int input;
while (isRunning) {
System.out.println("insert a number:");
try {
scanner = new Scanner(System.in);
input = scanner.nextInt();
inputToString(input);
isRunning = false;
} catch (InputMismatchException e) {
System.out.println("input musi byt cele cislo");
isRunning = true;
}
}
}
public static void inputToString(int input) {
System.out.println(input);
}
this works
I am having trouble with entering non-integers into an integer field. I am only taking precautions so that if another person uses/works on my program they don't get this InputMismatchException.
When I enter a non-digit character into the input variable, I get the above error. Is there any way to compensate for this like one could do for a NullPointerException when it comes to strings?
This code is redacted just to include the relevant portions causing the problem.
import java.util.Scanner;
class MyWorld {
public static void main(String[] args) {
Scanner user_input = new Scanner(System.in);
int input = 0;
System.out.println("What is your age? : ");
input = user_input.nextInt();
System.out.println("You are: " +input+ " years old");
}
}
You can use an if statement to check if user_input hasNextInt(). If the input is an integer, then set input equal to user_input.nextInt(). Otherwise, display a message stating that the input is invalid. This should prevent exceptions.
System.out.println("What is your age? : ");
if(user_input.hasNextInt()) {
input = user_input.nextInt();
}
else {
System.out.println("That is not an integer.");
}
Here is some more information about hasNextInt() from Javadocs.
On a side note, variable names in Java should follow the lowerMixedCase convention. For example, user_input should be changed to userInput.
You can add a try-catch block:
import java.util.Scanner;
class MyWorld {
public static void main(String[] args) {
Scanner user_input = new Scanner(System.in);
int input = 0;
System.out.println("What is your age? : ");
try{
input = user_input.nextInt();
}catch(InputMisMatchException ex)
System.out.println("An error ocurred");
}
System.out.println("You are: " +input+ " years old");
}
}
If you want to provide the user to enter another int you can create a boolean variable and make a do-while loop to repeat it. As follows:
boolean end = false;
//code
do
{
try{
input = user_input.nextInt();
end = true;
}catch(InputMisMatchException ex)
System.out.println("An error ocurred");
end = false;
System.out.println("Try again");
input.nextLine();
}
}while(end == false);
This is a try-catch block. You need to use this if you want to be sure of not making the program-flow stop.
try {
input = user_input.nextInt();
}
catch (InputMismatchException exception) { //here you can catch that exception, so program will not stop
System.out.println("Integers only, please."); //this is a comment
scanner.nextLine(); //gives a possibility to try giving an input again
}
Test using hasNextInt().
Scanner user_input = new Scanner(System.in);
System.out.println("What is your age?");
if (user_input.hasNextInt()) {
int input = user_input.nextInt();
System.out.println("You are " + input + " years old");
} else {
System.out.println("You are a baby");
}
Use Scanner's next() method to get data instead of using nextInt(). Then parse it to integer using int input = Integer.parseInt(inputString);
parseInt() method throws NumberFormatException if it is not int, which you can handle accordingly.
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();
}
}
This question already has answers here:
Determine if a String is an Integer in Java [duplicate]
(9 answers)
user input check int only
(4 answers)
Closed 9 years ago.
I'm trying to use while loop to ask the user to reenter if the input is not an integer
for eg. input being any float or string
int input;
Scanner scan = new Scanner (System.in);
System.out.print ("Enter the number of miles: ");
input = scan.nextInt();
while (input == int) // This is where the problem is
{
System.out.print("Invalid input. Please reenter: ");
input = scan.nextInt();
}
I can't think of a way to do this. I've just been introduced to java
The issue here is that scan.nextInt() will actually throw an InputMismatchException if the input cannot be parsed as an int.
Consider this as an alternative:
Scanner scan = new Scanner(System.in);
System.out.print("Enter the number of miles: ");
int input;
while (true) {
try {
input = scan.nextInt();
break;
}
catch (InputMismatchException e) {
System.out.print("Invalid input. Please reenter: ");
scan.nextLine();
}
}
System.out.println("you entered: " + input);
The javadocs say that the method throws a InputMismatchException if the input doesn;t match the Integer regex. Perhaps this is what you need?
So...
int input = -1;
while(input < 0) {
try {
input = scan.nextInt();
} catch(InputMismatchException e) {
System.out.print("Invalid input. Please reenter: ");
}
}
as an example.