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.
Related
I'm confused while using an Java program I created.
public static void main(String[] args) {
Scanner scanner1 = new Scanner(System.in);
int input1 = 0;
boolean Input1Real = false;
System.out.print("Your first input integer? ");
while (!Input1Real) {
String line = scanner1.nextLine();
try {
input1 = Integer.parseInt(line);
Input1Real = true;
}
catch (NumberFormatException e) {
System.out.println("Use an integer! Try again!");
System.out.print("Your first input integer? ");
}
}
System.out.println("Your first input is " + input1);
}
Initially, when a user Ctrl+D during the input, it will promptly end the program and display an error in the form of this,
Your first input integer? ^D
Class transformation time: 0.0073103s for 244 classes or 2.9960245901639343E-5s per class
Exception in thread "main" java.util.NoSuchElementException: No line found
at java.base/java.util.Scanner.nextLine(Scanner.java:1651);
at Playground.Test1.main(Test1.java:13)
Doing a bit of research I note that Ctrl+D terminates the input of sort. Therefore, I tried add few more lines to my codes to prevent the error from appearing again and instead printing a simple "Console has been terminated successfully!" and as far as my skills can go.
public static void main(String[] args) {
Scanner scanner1 = new Scanner(System.in);
int input1 = 0;
boolean Input1Real = false;
System.out.print("Your first input integer? ");
while (!Input1Real) {
String line = scanner1.nextLine();
try {
try {
input1 = Integer.parseInt(line);
Input1Real = true;
}
catch (NumberFormatException e) {
System.out.println("Use an integer! Try again!");
System.out.print("Your first input integer? ");
}
}
catch (NoSuchElementException e) {
System.out.println("Console has been terminated successfully!");
}
}
System.out.println("Your first input is " + input1);
}
In the end, I still got the same error.
Got it!, the code hasNext() will ensure that the error will not appear. This method is to check whether there is another line in the input of the scanner and to check if its filled or empty. I am also using null to check my statement after passing the loop so the program stops if the input value is still null while keeping the function of Ctrl+D.
public static void main(String[] args) {
Integer input1 = null;
System.out.println("Your first input integer? ");
Scanner scanner1 = new Scanner(System.in);
while(scanner1.hasNextLine()) {
String line = scanner1.nextLine();
try {
input1 = Integer.parseInt(line);
break;
}
catch (NumberFormatException e) {
System.out.println("Use an integer! Try again!");
System.out.println("Your first input integer? ");
}
}
if (input1 == null) {
System.out.println("Console has been terminated successfully!");
System.exit(0);
}
System.out.println(input1);
}
This solution is not prefect of course but I would appreciate if there were much simpler options.
I am trying to create a method for error checking input using a try and catch statement with the catch InputMismatch. It is not working can anyone advise why my return statement is not returning the input from the user and storing it in the integer which is called in the main class(call statement is not included).
System.out.print("Please enter the percent achieved for " + sName + ": %");
PromptErrorCheckingPercent(iPercent);
public static int PromptErrorCheckingPercent(int test){
Scanner keyboard = new Scanner(System.in);
boolean valid = false;
while (!valid)
{
try
{
test = keyboard.nextInt();
valid = true;
}
catch(InputMismatchException e)
{
System.out.println("Invalid Input. Please enter a valid integer");
keyboard.nextLine(); //nextLine for a reason
}
}
return test;
What you need is a recursion when InputMismatchException is thrown instead of keyboard.nextLine(); //nextLine for a reason. Moreover, you do not need to pass the value in the argument/ parameter of promptErrorCheckingPercent() method because what you are trying is not allowed in JAVA.
Probably you were thinking passing that variable will help you to store the data in that variable. It is a feature of C-language that supports Pointer (address of the variable) but for the sake of security it is not allowed in JAVA.
You can look into the code below to check if it can solve your problem.
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.println("Please enter the name:");
String sName = keyboard.nextLine();
System.out.print("Please enter the percent achieved for " + sName + ": %");
int iPercent = promptErrorCheckingPercent();
}
public static int promptErrorCheckingPercent() {
Scanner keyboard = new Scanner(System.in);
boolean valid = false;
int test = -1;
while (!valid) {
try {
test = keyboard.nextInt();
valid = true;
} catch(InputMismatchException e) {
System.out.println("Invalid Input. Please enter a valid integer");
return promptErrorCheckingPercent(); // Recursion by calling the method itself
}
}
return test;
}
}
I want the program to not crash when the user inputs anything other than a number. For example, if someone enters random letters, it should display a message saying, "input not valid, please enter a valid integer". Then prompting them if they would want to continue or not.
public static void main(String[] args) throws IOException {
BufferedWriter out = new BufferedWriter(new FileWriter("outDataFile.txt"));
Scanner input=new Scanner(System.in);
int choice = 0;
String repeat;
//Loop repeats program until user quits
do
{
//Loop repeats until a valid input is found
do
{
//Asks for number and reads value
System.out.print("\nEnter an integer and press <Enter> ");
choice = input.nextInt();
//Prints error if invalid number
if(choice <= 0)
System.out.println("Invalid Input.");
There are multiple ways to achieve that:
First is to catch the exception thrown by the Scanner and flag the loop to carry on when an exception is caught. This is not a good practise since the exception thrown by the Scanner, InputMismatchException, is an unchecked exception. Meaning the cause of this exception can be easily caught by an if/else statement.
In your case, you should try to receive the input as a String, then validate the input if it looks like a number:
Loop per character approach:
String string = scanner.nextLine();
for (int i = 0; i < string; i++) {
char ch = string.charAt(i);
if (!Character.isDigit(ch)) {
System.out.println("Input is not a number");
break; // stop the for-loop
}
}
int input = Integer.parseInt(string);
RegEx approach:
String numericRegex = "[0-9]+";
String string = scanner.nextLine();
if (!string.matches(numericRegex)) {
System.out.println("Input is not a number");
}
int input = Integer.parseInt(string);
These are popular approach to your problem, it is now up to you on how will you control your loops to repeat when an invalid input is encountered.
Use a simple try catch that will catch and a simple recursive method as such:
import java.util.InputMismatchException;
import java.util.Scanner;
public class Test
{
public static void main(String[] args)
{
System.out.println(getUserInput());
}
private static int getUserInput()
{
int choice = 0;
Scanner input = new Scanner(System.in);
System.out.println("Enter a value");
try
{
choice = input.nextInt();
} catch (InputMismatchException exception)
{
System.out.println("Invalid input. Please enter a numeric value");
getUserInput();
}
return choice;
}
}
boolean loop = false;
double numberOfStudents;
System.out.print("Enter a number: ");
if ((scnr.nextLine().trim().isEmpty()) ) {
loop = true;
}
while (loop) {
System.out.println("Enter a number");
if (scnr.hasNextDouble() ){
System.out.println("Loop has stopped");
numberOfStudents = scnr.nextDouble();
loop = false;
}
}
System.out.println("You're outside the loop!");
I'm trying to get the program to say "Enter a number" until the user has entered an actual number (no white spaces or letters or signs). When the user has entered a number, it sets numberOfStudents equal to that number and breaks out of the loop.
But if you hit enter twice, it doesn't iterate. It only displays "Enter a number" once.
What is wrong with the loop logic? Why isn't it looping until valid input is taken?
For the actual answer to your question of "Why doesn't 'Enter a number' display more than once?" see Tom's comment (update: Tom's answer).
I've rewritten your loop in a way which preserves your code, but also makes it a little easier to handle format exceptions (though at the risk of silently swallowing an exception -- should be acceptable for this use case).
Can be up to you to use this design, here is an SO post on why empty catch blocks can be a bad practice.
public static void main(String args[])
{
boolean loop = true;
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
while(loop){
System.out.print("Enter a number: ");
String input = scnr.nextLine();
try{
numberOfStudents = Double.parseDouble(input);
loop = false;
}catch(NumberFormatException e){
}
}
System.out.println("You're outside the loop!");
}
Output:
Enter a number:
Enter a number:
Enter a number:
Enter a number: 50
You're outside the loop!
First of all: Since you're reading from System.in a call to the input stream will block until the user entered a valid token.
So let's check first scan using your scnr variable:
scnr.nextLine()
nextLine() reads everything til the next line delimiter. So if you just press return, then it will successfully read it and will perform the next stuff.
The next call is:
scnr.hasNextDouble()
This call expects a "real" token and ignores white spaces, except as a delimiter between tokens. So if you just press return again it doesn't actually read that input. So it still waits for more (for the first token). That is why it stucks in your loop and you won't get another "Enter a number" output.
You can fix that by either enter a real token, like a number, or by changing the loop like trobbins said.
I hope you now understand your program flow a bit more :).
While trobbins code basically solves your problem, it's bad practice to use exceptions for flow control.
I used a small regexp to check if the value is a number. But this example is not complete, it will still crash it the user enters for example two decimal points. So you would need to create a proper number check or just use integers where the check is much easier.
Someone in the comments pointed out that people may want to enter scientific notation like 5e10, so this would also be another case to check for. If this is just some code you need as a proof of concept or something quick and dirty, you can go with the exception handling method but in production code you should avoid using exceptions this way.
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
while(true) {
System.out.print("Enter a number: ");
String input = scnr.nextLine().trim();
if(input.matches("^[0-9\\.]{1,}$")) {
System.out.println("Loop has stopped");
numberOfStudents = Double.parseDouble(input);
break;
}
}
System.out.println("You're outside the loop!");
The following code should help you:
double numberOfStudents = 0;
Scanner scnr = new Scanner(System.in);
boolean readValue = false; //Check if the valid input is received
boolean shouldAskForNumber = true; //Need to ask for number again? Case for Enter
do {
if (shouldAskForNumber) {
System.out.print("Enter a number:");
shouldAskForNumber = false;
}
if (scnr.hasNextDouble()) {
numberOfStudents = scnr.nextDouble();
readValue = true;
} else {
String token = scnr.next();
if (!"".equals(token.trim())) { //Check for Enter or space
shouldAskForNumber = true;
}
}
} while (!readValue);
System.out.printf("Value read is %.0f\n", numberOfStudents);
System.out.println("You're outside the loop!");
Update
Understood the following statement in question different way:
But if you hit enter twice, it doesn't loop back. It only displays
"Enter a number" once.
The code is set to print "Enter a number" only once if the user hits RETURN/ENTER or enters space character. You may remove the special check and use the code if needed.
import java.util.Scanner;
public class Testing {
public static boolean checkInt(String s)
{
try
{
Integer.parseInt(s);
return true;
} catch (NumberFormatException ex)
{
return false;
}
}
public static void main(String[] args) {
boolean loop = false;
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
String input = "";
while (!(checkInt(input))) {
System.out.println("Enter a number");
input = scnr.nextLine();
}
numberOfStudents = Integer.parseInt(input);
System.out.println("Number of students: " + numberOfStudents );
}
}
//this code is working fine, if you want you check it out.
//In your code your taking another input if the first is an int/double; if the first input is not a number then you have mentioned to take input again..
Use a debugger to see what the code is actually doing. Here's a guide on debugging in Eclipse. After you have finished debugging your code, you will probably know what the problem is.
Below code will help you
boolean loop = true;
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
System.out.print("Enter a number: ");
String input = scnr.nextLine();
while(!scnr.hasNextDouble()){
System.out.print("Enter a number: ");
try{
numberOfStudents = Double.parseDouble(input);
break;
}catch(NumberFormatException e){
}
input = scnr.nextLine();
}
System.out.println("You're outside the loop!");
The following code is working,
boolean loop = true;
double numberOfStudents;
Scanner scnr=new Scanner(System.in);
while(loop) {
System.out.println("Enter a number");
if ((scnr.nextLine().trim().isEmpty()) ) {
loop = true;
}
if (scnr.hasNextDouble() ){
System.out.println("Loop has stopped");
numberOfStudents = scnr.nextDouble();
loop = false;
}
}
System.out.println("You're outside the loop!");
The output is,
run:
Enter a number
hj
po
Enter a number
lhf
Enter a number
o
Enter a number
p
Enter a number
a
Enter a number
34
Loop has stopped
You're outside the loop!
You have to scan the next line if you want to get more values form the scanner again. The code should be like:
while (loop) {
System.out.println("Enter a number");
if(!(scnr.nextLine().trim().isEmpty())){
if (scnr.hasNextDouble() ){
System.out.println("Loop has stopped");
numberOfStudents = scnr.nextDouble();
loop = false;
}
}
}
Goal:
If the user enters a non-numeric number, make the loop run again.
Also is there another (more efficient) way of writing the numeric inputs?
public static void user_input (){
int input;
input = fgetc (System.in);
while (input != '\n'){
System.out.println("Please enter a number: ");
if (input == '0' == '1' ..... '9'){
//Execute some code
}
else {
System.out.println("Error Please Try Again");
//Repeat While loop
}
}
}
EDIT
I need the while loop condition. Simply asking, how do you repeat the while loop? Also no scanner methods.
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:
public static void user_input() {
Scanner sc = new Scanner(System.in);
while (true) {
System.out.println("Enter a 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");
}
}
}
Output
Enter a number.
w
Input is not a number, continue
Enter a number.
3
Correct input, exit
Try this one
System.out.println("Please enter a number: ");
Scanner userInput = new Scanner(System.in);
while(!userInput.hasNextInt()) {
System.out.println("Invalid input. Please enter again");
userInput = new Scanner(System.in);
}
System.out.println("Input is correct : " + userInput.nextInt());
How about this
public static void processInput() {
System.out.println("Enter only numeric: ");
Scanner scannerInput;
while (true) {
scannerInput = new Scanner(System.in);
if (scannerInput.hasNextInt()) {
System.out.println("Entered numeric is " + scannerInput.nextInt());
break;
} else {
System.out.println("Error Please Try Again");
}
}
}