I am getting an error when I am entering string value in integer variable. I want to know to handle the exception, as my program give a indication that enter your value again rather it stop and gave exceptional handling error.
System.out.println("Please enter a number");
Scanner obj1=new Scanner(System.in);
int a=obj1.nextInt(); //I am entering string value here
System.out.println(a);
Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Scanner.java:864)
at java.util.Scanner.next(Scanner.java:1485)
at java.util.Scanner.nextInt(Scanner.java:2117)
at java.util.Scanner.nextInt(Scanner.java:2076)
at javaapplication13.JavaApplication13.main(JavaApplication13.java:23)
It is better to preventing throwing exception instead of handling it since creating exception object may be quite expensive. In case of Scanner class your code can look like:
Scanner scanner = new Scanner(System.in);
System.out.println("Please enter a number:");
while(!scanner.hasNextInt()){//while user is not providing valid int
//inform him about it
System.out.println("that was not valid integer, please try again: ");
//and consume invalid value (maybe even entire line)
scanner.nextLine();
}
//since we are out of loop it must mean that user entered valid int, lets store it
int a = scanner.nextInt();
System.out.println(a);
You can also wrap this code in some nice utility method in which you can even add support to maximal number of tries. For that kind of code you can use as result Optional<Integer>. This will allow us clear info if user
provided proper value, and we will return Optional with this value
or if he failed to do so, in which case we can return empty Optional.
which is cleanest way of handling results which may not exist.
Without Optional we would either have to:
return some default value like:
-1 but in that case we don't know if -1 represents invalid input, or if user actually provided -1 which means it may be proper result
null (if we change return type to Integer) but this way we are making our programmer who will use our method to also check if(result != null) which may not be that obvious for many programmers (using Optional gives us clear idea that result may not exist so programmer knows that he should check this possibility).
throw exception, but since reason why we are creating our method is to avoid creating and handling exception it would be kind of stupid.
So your method can look like
//-1 represents infinite number of tries
public static Optional<Integer> readInt(Scanner sc, String msg, int tries){
System.out.println(msg);
int counter = 0;
//while user still has more tries but is providing invalid int
while((tries == -1 || counter < tries) && !sc.hasNextInt()){
//inform him about wrong data
if (tries == -1 || ++counter < tries)
System.out.println("That was not valid integer. Please try again:");
else
System.out.println("That was not valid integer.");
//and consume invalid value (maybe even entire line)
sc.nextLine();
}
// since we are out of loop it must mean that user entered valid int
// or run out of tries
if (tries == -1 || counter < tries)
return Optional.of(sc.nextInt());
else
return Optional.empty();
}
and its usage can look like
Scanner scanner = new Scanner(System.in);
Optional<Integer> result = readInt(scanner, "Please enter a number:", 1);
if(result.isPresent()){
System.out.println(result.get());
}else{
System.out.println("User failed to give valid integer. Program will terminate");
}
Use this
System.out.println("Please enter a number");
Scanner obj1=new Scanner(System.in);
if(obj1.hasNextInt())
{
int a=obj1.nextInt(); //I am entering string value here
System.out.println(a);
}
Put the code in the try-catch block to handle the InputMismatchException
System.out.println("Please enter a number");
Scanner obj1=new Scanner(System.in);
try {
int a=obj1.nextInt();
}
catch(InputMismatchException ime) {
/*show error message for incorrect input*/
}
catch(Exception e) {
/*Show error message*/
}
System.out.println(a);
Hope, this solves the problem.
You can try this...
boolean i=true;
while(i){
try{
System.out.println("Please enter a number");
Scanner obj1=new Scanner(System.in);
int a=obj1.nextInt(); //I am entering string value here
System.out.println(a);
i=false;
}catch(InputMismatchException e){
System.out.println("Invalid number, please enter again");
}
}
Related
If the user enter a non-numeric value for example "Hello", I want to give the user a new chance to enter a numeric value until he succeeds.
Running the code below, the text "This is not a numeric value. Enter a numeric value: ", will show in the console, if the user enters a non-numeric value. However, the user will not be able to enter a new value. Because the error message "Exception in thread..." will show and stop the program.
How do I solve this?
Edit!
New code (This code works, except it stops the program completely. Every method called afterwards, won't run.) Edit2! It did not work!
int number;
do {
try {
System.out.println("Enter a numeric value: ");
number = Integer.parseInt(s.nextLine());
if (s.nextInt() != (int)number) {
throw new NumberFormatException();
}
break;
}
catch (NumberFormatException e) {
System.out.println("This is not a numeric value. Enter a numeric value: ");
}
} while (true)
Old code
int number;
try {
Scanner s = new Scanner(System.in);
System.out.println("Enter a numeric value: ");
number = s.nextInt();
if (number != (int)number) {
throw new Exception();
}
}
catch(Exception e) {
do {
System.out.println("This is not a numeric value. Enter a numeric value: ");
number = s.nextInt();
} while (number != (int)number);
}
Use NumberFormatException instead of Base Exception Class
int number;
Scanner s = new Scanner(System.in);
System.out.println("Enter a numeric value: ");
do {
try {
number = Integer.parseInt(s.nextLine());
//Do your work
break;
} catch (NumberFormatException e) {
System.out.println("This is not a numeric value. Enter a numeric value: ");
}
}
while (true);
}
Please try below given code:
int number;
Scanner s = new Scanner (System.in);
System.out.println ("Enter a numeric value: ");
do
{
try
{
number = Integer.parseInt (s.nextLine ());
//Do your work
break;
}
catch (Exception e)
{
System.out.println
("This is not a numeric value. Enter a numeric value: ");
}
}
while (true);
}
It sounds like you want to loop forever until the user enters a valid integer. A common way to express "loop forever" is with while (true) {...}. This is an "infinite loop." The only way to exit this loop is with a break statement or a return statement (or by throwing an exception, but in this case you want to catch the exception and continue looping).
One problem with using scanner.nextInt() is that if the value is not an integer, the value will remain in the scanner's buffer. If you keep calling nextInt(), you'll just keep getting InputMismatchExceptions since the scanner will keep trying to interpret the same bad input as an integer, over and over again.
One way around that problem is to use scanner.nextLine() to read the value as a String, and then use Integer.parseInt(String) to convert the String to an int. If the conversion fails, Integer.parseInt(String) throws a NumberFormatException. You can handle this exception by catching it.
Here's a little function that loops forever until the user enters a value that can be parsed as an int:
public static int promptForInt(Scanner scanner) {
while (true) {
System.out.println("Enter a numeric value:");
try {
return Integer.parseInt(scanner.nextLine());
} catch (NumberFormatException e) {
System.out.print("This is not a numeric value. ");
}
}
}
You can call the method like this:
Scanner scanner = new Scanner(System.in);
int number = promptForInt(scanner);
I don't think you truly understand where the exception is coming from. The line throw new Exception(); never runs in your code. This is because your if statement says if (number != (int) number), which will never be true because the variable number is already of the type int, so you are basically saying if (number != number), which can never be true. If you take a look at Scanner's nextInt() method, you'll notice that it actually throws InputMismatchException. That is the exception you are catching with your try-catch, but it is not the exception that is causing the error message. So where is that exception coming from? The line number = s.nextInt(); is not compilable code. You defined s inside the try block, so therefore s only exists inside the try block. You get an exception when you try to access it in the catch block because s is not a variable there; it does not exist as far as the compiler is concerned. The fix is pretty simple; just move the declaration for s to outside the try block, and put the while loop around the try-catch block. Also remember to consume the line separators in the input stream using s.nextLine() every time you try to read an int. Here's an example of how I would do it:
int number;
Scanner s = new Scanner(System.in); //declare s outside of the try block
System.out.println("Enter a numeric value: ");
while (true) { //while loop goes around the try-catch block
try {
number = s.nextInt(); //this could throw InputMismatchException
break; //if no exception is thrown, break out of the loop
} catch (InputMismatchException e) { //if the exception is thrown
s.nextLine(); //consume the line separator character(s)
System.out.println("This is not a numeric value. Enter a numeric value: "); //prompt the user for another value
//this will then go back to the top of the try block again, because we never broke out of the while loop
}
}
s.nextLine(); //consume the line separator character(s)
Try the following, if you take your input as a string you can parseInt to check if its an integer value
String number = null;
Scanner s = new Scanner(System.in);
System.out.println("Enter a numeric value: ");
number = s.next();
try {
// checking valid integer using parseInt() method
Integer.parseInt(number);
} catch (NumberFormatException e) {
System.out.println("This is not a numeric value. Enter a numeric value:");
number = s.next();
// put this line here to prove it accepted the second attempt
System.out.println("Your entered integer is:" + number);
}
s.close();
}
}
I pretty new to java programming so i was wondering if there is a way to use the condition of a while loop to stop an invalid value being used.
I am writing a program that prompts the user to enter an identification number as an integer then uses a scanner to store that value.
Just wanted to know if this is possible to put something in the condition of the for loop that prints an error message if the enter something like a string, double or char so i dont get the Input Mismatch Exception.
like this:
identification = userId(in); //scanner
while (identification (is not an integer)){
System.out.println("Invalid Value, Please enter an integer");
identification = userId(in);
Even better, you can write:
while ((identification = userId(in)) < 0) {
System.out.println("blah ...");
}
The assumption is that the userIn method returns some negative value if the input is not an integer. You can make the invalid return value whatever you want as long as it is not something that is a valid input.
Some people don't like this style because it has gone out of fashion and they are not used to it; however, it used to be common in C programming and there is nothing about it that is implicitly unclear or bad.
This should do what you are asking. It is basically a while loop that waits until the next input is an integer before continuing the code. The important part is making sure to use in.next() inside the while loop instead of in.nextInt() because the values could be anything.
Scanner in = new Scanner(System.in);
System.out.print("Enter an integer: ");
while (!(in.hasNextInt()))
{
System.out.print("Integer not entered, please enter an integer: ");
in.next();
}
int value = in.nextInt();
System.out.println("The int was " + value);
in.close();
boolean acceptPcode=true;
boolean acceptQty=false;
int Qty=0;
List<Integer> purchasedProdQty=new ArrayList<>();
while(acceptPcode==true && acceptQty==false){
do{
try{
System.out.print("Enter Qty: ");
Qty=sc.nextInt();
acceptQty=true;
}catch(InputMismatchException ime){
System.out.println("Invalid quantity please enter a number!");
acceptQty=false;
}
if(acceptQty==true)
purchaseProdQty.add(Qty);
}while(acceptQty==false);
}
my question is that when i enter a letter it goes in an infinity loop and it doesn't prompt the user to enter a quantity ....which is
Enter Qty: Invalid quantity please enter a number!
Enter Qty: Invalid quantity please enter a number!
Enter Qty: Invalid quantity please enter a number!......
You forgot to read the \n (or \r\n) characters that are from the next line. In your current code, the scanner is waiting for an int input, bu the current next input is this break line char. Just add sc.nextLine() in your code to consume the break line char:
Qty=sc.nextInt();
sc.nextLine();
acceptQty=true;
From what I can gather it seems your scanner(sc) is throwing an exception. This causes acceptQty to constantly be false keeping you stuck in your inner do-while loop.
You need to consume any illegal characters in the exception block otherwise they won't be consumed by the Scanner#nextInt method call causing the loop to repeat itself indefinitely:
} catch(InputMismatchException ime) {
System.out.println
("Invalid quantity: " + sc.nextLine() + " please enter a number ");
...
}
You are getting exception while reading out of sc and so it always go into infinity loop. Can you paste what's the value assiged in sc?
I believe you're doing this all wrong. Your method of validation is very obscure and can be simplified. Suppose you have the following method:
public int readNumber(final String prompt, final Scanner scanner){
System.out.println(prompt);
try{
return scanner.nextInt();
}catch(Exception ex){
System.err.println("Enter a valid number");
return readNumber(prompt, scanner);
}
}
This method will print out the prompt (the first argument) and read input from the provided Scanner (the second argument). If the user enters something that can't be parsed as an int, it will invoke the same method (recursion).
Take out both of your loops and when you want to read an int from your Scanner, do something like:
int value = readNumber("Enter a quantity", sc);
You know for sure that Integer.MAX_VALUE >= value >= Integer.MIN_VALUE
I wrote a program to determine a prime number and am required to use 'Q' as the quit function to get out of the program. Below is the code i have written. I am wondering how I would modify method of quitting the program to make it exit correctly.
import java.util.Scanner;
public class Practice
{
public static void main(String[]args)
{
Scanner keyboard = new Scanner(System.in);
int number, i=2;
String quit;
boolean prime = true;
System.out.println("Please enter a number to determine if it is prime, to quit please enter Q.");
number = keyboard.nextInt();
quit = keyboard.nextLine();
for (i=0; i<number;i++)
{
if ((number%2==0) || (number==1) || (number%3==0))
{
prime = false;
}
}
if (prime==false)
{
System.out.println(number+" is not a prime number.");
}
else if(prime==true)
System.out.println(number+" is a prime number.");
if (quit.charAt(0)=='Q')
{ System.exit(0);}
and my output upon entering 'Q' is:
----jGRASP exec: java Practice
Please enter a number to determine if it is prime, to quit please enter Q.
Q
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.nextInt(Scanner.java:2160)
at java.util.Scanner.nextInt(Scanner.java:2119)
at Practice.main(Practice.java:15)
----jGRASP wedge2: exit code for process is 1.
----jGRASP: operation complete.
}
}
When the input through the console is given as "Q", the value of type String is getting assigned to a variable of type int, which is incorrect.
number = keyboard.nextInt();
The type of in the input needs to be checked, before it can be assigned to a int variable.
You have:
number = keyboard.nextInt();
quit = keyboard.nextLine();
Think for a moment about what actually happens here. 'Q' is not a number and so nextInt(), as documented, throws an InputMismatchException. Your use of nextLine() doesn't make much sense following that.
You have a couple of options:
Use Scanner.hasNextInt() to determine if the next token truly is a well-formed integer. If not, read it as a string and check it.
Use Scanner.next() always, and check if it is "Q" before parsing it with Integer.parseInt() (being prepared to handle errors there as well).
The Scanner.next* functions don't implicitly skip invalid values. The Scanner is only doing what you tell it to do, and if you tell it to grab an integer but the input is not an integer, it rightfully complains. It is up to you to use the Scanner appropriately and check your possible input cases.
You should read in the numbers as Strings with keyboard.nextLine();
and check first if the it contains 'Q'. If it does, quit, else do, number = Integer.parseInt(yourVariableHere); and go from there
Try this instead:
String input;
int number = 1;
int i=2;
Then do this:
System.out.println("Please enter a number to determine if it is prime, "
+ "\nto quit, please enter Q.");
input = keyboard.nextLine();
if (input.charAt(0)=='Q')
{
System.exit(0);
}
else
{
try
{
number = Integer.parseInt(input);
}
catch(NumberFormatException nfe)
{
System.out.println("Please enter a number or Q!");
}
}
Basic problem here.. I will start off by asking that you please not respond with any code, as that likely will only confuse me further (programming noob). I am looking for a clear explanation on how to solve this issue that I'm having.
I have a scanner that reads input from the user. The user is prompted to enter an int value between 1 to 150 (whole numbers only). I obtain the value as follows:
Scanner scan = new Scanner(System.in);
int input = scan.nextInt();
And continue on with my program, and everything works fine.
Unfortunately, the code isn't exactly bulletproof, since any input that is not an integer can break it (letters, symbols, etc).
How can I make the code more robust, where it would verify that only an int was entered?
These are the results I'm hoping for:
Lets say the input was:
23 -> valid
fx -> display an error message, ask the user for input again (a while loop would do..)
7w -> error, again
3.7 -> error
$$ -> error
etc
Scanner.hasNextInt() returns true if the next token is a number, returns false otherwise.
In this example, I call hasNextInt(). If it returns true, I go past the while and set the input; if it returns false, then I discard the input (scanner.next();) and repeat.
Scanner scan = new Scanner(System.in);
while(!scan.hasNextInt()) {
scan.next();
}
int input = scan.nextInt();
Here's a simple example with prompts and comments.
Scanner scan = new Scanner(System.in);
System.out.print("Enter an integer: "); // Initial prompt for input
// Repeat until next item is an integer
while (!scan.hasNextInt())
{
scan.next(); // Read and discard offending non-int input
System.out.print("Please enter an integer: "); // Re-prompt
}
// At this point in the code, the user has entered an integer
int input = scan.nextInt(); // Get the integer
// And now you can use the input variable.
Use scan.hasNextInt() to make sure the next input is an int.
I have written an example that ensures that the program will continue only if a number and not an invalid value is entered. Do not worry, I added the desired explanation.
The program asks the user to input a number. A loop ensures that the processing will not go on until a valid number is entered. Before that I have defined a variable "inputAccepted" that has false as default value. If he enters a number, the variable "inputAccepted" is set to true and the program leaves the loop. But if he enters something else than a number, an exception is thrown right in this moment, and the line that sets the variable "inputAccepted" to true will not be executed. Instead a message will be printed out that tells the user that his input is not valid. Since "inputAccepted" could not be set to true, the loop will do the same stuff again until the string can be converted to a number.
You can test the program here.
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean inputAccepted = false;
while (!inputAccepted) {
try {
System.out.print("Please enter a number: ");
Integer.valueOf(input.nextLine());
inputAccepted = true;
} catch (NumberFormatException e) {
System.out.println("Not a valid number.");
}
}
System.out.println("Thank you!");
}
}
Just get "anything" and parse it:
Scanner scan = new Scanner(System.in);
Integer number = null;
while (number == null) {
try {
number = Integer.parseInt(scan.next());
} catch (NumberParseException e) {
System.out.println("bad input: " + input);
}
}
Without any code and just in English, I'd say there's two things you have to test or look out for. First that the input is an int, second that the int is within the correct range.
In terms of pseudocode, the first thing to do is make sure it's an int. Declaring an int named "input", I would put a try / catch block, where you try to scan in the user input as an int, with parseInt(). If the try part fails, you know it's not an int and can return an error message.
Then, now that you know that "input" is an int, you can test whether it is less than 1 or more than 150, and return an error message if so!
public class Sample {
/**
* author CLRZ
*/
public static void main(String[] args) {
int a; // variable
Scanner in = new Scanner(System.in); // scans your input
System.out.println("Enter your number's choice:");
int sem1 = in.nextInt(); // reads next integer
if (sem1 == 1) // conditioned if your choice number is equal to 1
System.out.println("Hello World1"); // output wil be Hello World
int b;
System.out.println("Enter your number's choice:");
int sem2 = in.nextInt();
if (sem2 == 2)
System.out.println("Hello World2");
int c;
System.out.println("Enter your number's choice:");
int sem3 = in.nextInt();
if (sem3 == 3)
System.out.println("Hello World3");
}
}