I am just trying to get code to work where the code asks again for an answer, if text or a symbol is entered, instead of a required integer:
import java.util.Scanner;
class timecalc {
int hrs = 0;
int min = 0;
static int hourflag = 0;
static int minflag = 0;
Scanner sc = new Scanner(System.in);
public int getHours() {
try {
hourflag = hourflag + 1;
if (hourflag > 1) {
System.out.println("Invalid month Please enter hours again:");
}
System.out.println("Enter month:");
return hrs = sc.nextInt();
} catch (InputMisMAtchException e) {
System.out.println("entered invalid input " + e);
}
}
Have reviewed answers already given but cant get a workable solution
Any ideas?
I won't give you the entire code, but just a hint or psuedo-code. As an exercise you can implement it as per your requirement.
System.out.println("Enter month:");
while (true) {
try {
int min = sc.nextInt();
break;
} catch (InputMismatchException ex) {
System.err.println("Invalid input, please enter again");
sc.nextLine(); // <----- advance the scanner
}
}
Here the logic is to loop until we get the right input. If it is an invalid input, the loop never breaks.
Also as a side-note, I would recommend you to create just one method to fetch correct inputs and call it respectively from other methods. Rather than duplicating this logic everywhere.
Related
This question already has answers here:
How to handle infinite loop caused by invalid input (InputMismatchException) using Scanner
(5 answers)
Closed last month.
So I'm building a program which takes ints from user input. I have what seems to be a very straightforward try/catch block which, if the user doesn't enter an int, should repeat the block until they do. Here's the relevant part of the code:
import java.util.InputMismatchException;
import java.util.Scanner;
public class Except {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean bError = true;
int n1 = 0, n2 = 0, nQuotient = 0;
do {
try {
System.out.println("Enter first num: ");
n1 = input.nextInt();
System.out.println("Enter second num: ");
n2 = input.nextInt();
nQuotient = n1/n2;
bError = false;
}
catch (Exception e) {
System.out.println("Error!");
}
} while (bError);
System.out.printf("%d/%d = %d",n1,n2, nQuotient);
}
}
If I enter a 0 for the second integer, then the try/catch does exactly what it's supposed to and makes me put it in again. But, if I have an InputMismatchException like by entering 5.5 for one of the numbers, it just shows my error message in an infinite loop. Why is this happening, and what can I do about it? (By the way, I have tried explicitly typing InputMismatchException as the argument to catch, but it didn't fix the problem.
You need to call next(); when you get the error. Also it is advisable to use hasNextInt()
catch (Exception e) {
System.out.println("Error!");
input.next();// Move to next other wise exception
}
Before reading integer value you need to make sure scanner has one. And you will not need exception handling like that.
Scanner scanner = new Scanner(System.in);
int n1 = 0, n2 = 0;
boolean bError = true;
while (bError) {
if (scanner.hasNextInt())
n1 = scanner.nextInt();
else {
scanner.next();
continue;
}
if (scanner.hasNextInt())
n2 = scanner.nextInt();
else {
scanner.next();
continue;
}
bError = false;
}
System.out.println(n1);
System.out.println(n2);
Javadoc of Scanner
When a scanner throws an InputMismatchException, the scanner will not pass the token that caused the exception, so that it may be retrieved or skipped via some other method.
YOu can also try the following
do {
try {
System.out.println("Enter first num: ");
n1 = Integer.parseInt(input.next());
System.out.println("Enter second num: ");
n2 = Integer.parseInt(input.next());
nQuotient = n1/n2;
bError = false;
}
catch (Exception e) {
System.out.println("Error!");
input.reset();
}
} while (bError);
another option is to define Scanner input = new Scanner(System.in); inside the try block, this will create a new object each time you need to re-enter the values.
To follow debobroto das's answer you can also put after
input.reset();
input.next();
I had the same problem and when I tried this. It completely fixed it.
As the bError = false statement is never reached in the try block, and the statement is struck to the input taken, it keeps printing the error in infinite loop.
Try using it this way by using hasNextInt()
catch (Exception e) {
System.out.println("Error!");
input.hasNextInt();
}
Or try using nextLine() coupled with Integer.parseInt() for taking input....
Scanner scan = new Scanner(System.in);
int num1 = Integer.parseInt(scan.nextLine());
int num2 = Integer.parseInt(scan.nextLine());
To complement the AmitD answer:
Just copy/pasted your program and had this output:
Error!
Enter first num:
.... infinite times ....
As you can see, the instruction:
n1 = input.nextInt();
Is continuously throwing the Exception when your double number is entered, and that's because your stream is not cleared. To fix it, follow the AmitD answer.
#Limp, your answer is right, just use .nextLine() while reading the input. Sample code:
do {
try {
System.out.println("Enter first num: ");
n1 = Integer.parseInt(input.nextLine());
System.out.println("Enter second num: ");
n2 = Integer.parseInt(input.nextLine());
nQuotient = n1 / n2;
bError = false;
} catch (Exception e) {
System.out.println("Error!");
}
} while (bError);
System.out.printf("%d/%d = %d", n1, n2, nQuotient);
Read the description of why this problem was caused in the link below. Look for the answer I posted for the detail in this thread.
Java Homework user input issue
Ok, I will briefly describe it. When you read input using nextInt(), you just read the number part but the ENDLINE character was still on the stream. That was the main cause. Now look at the code above, all I did is read the whole line and parse it , it still throws the exception and work the way you were expecting it to work. Rest of your code works fine.
The algorithm should take in 3 integers to an ArrayList. If the input is not an integer, then there should be a prompt. When I execute my code the catch clause is executed, but the program runs into a infinite loop. Could someone guide me into the right direction, I appreciate the help. :-D
package chapter_08;
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
public class IntegerList {
static List<Integer> numbers = new ArrayList<Integer>();
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int counter = 1;
int inputNum;
do {
System.out.print("Type " + counter + " integer: " );
try {
inputNum = input.nextInt();
numbers.add(inputNum);
counter += 1;
}
catch (Exception exc) {
System.out.println("invalid number");
}
} while (!(numbers.size() == 3));
}
}
That is because when the next int is read using nextInt() and it fails, the Scanner still contains the typed contents. Then, when re-entering the do-while loop, input.nextInt() tries to parse it again with the same contents.
You need to 'flush' the Scanner contents with nextLine():
catch (Exception exc) {
input.nextLine();
System.out.println("invalid number");
}
Notes:
You can remove the counter variable, because you're not using it. Otherwise, you could replace counter += 1 by counter++.
You can replace while (!(numbers.size() == 3)) with while (numbers.size() != 3), or even better: while (numbers.size() < 3).
When catching exceptions, you should be as specific as possible, unless you have a very good reason to do otherwise. Exception should be replaced by InputMismatchException in your case.
If inputNum = input.nextInt(); cannot be fit into an int and a InputMismatchException is raised, the input of the Scanner is not consumed.
So after the catch, it loops and it goes again here :
inputNum = input.nextInt();
with exactly the same content in the input.
So you should execute input.nextLine(); in the catch statement to discard the current input and allow a new input from the user.
Besides it makes more sense to catch InputMismatchException rather than Exception as other exception with no relation with a mismatch could occur and it would not be useful to display to the user "invalid number " if it is not the issue :
catch (InputMismatchException e){
System.out.println("invalid number ");
input.nextLine();
}
You should to use a break; in your catch(){} like so :
try {
inputNum = input.nextInt();
numbers.add(inputNum);
counter += 1;
} catch (Exception e) {
System.out.println("invalid number ");
break;
}
So if one input is not correct break your loop.
try changing
inputNum = input.nextInt();
to
String inputText=input.next();
inputNum = Integer.valueOf(inputText);
it works perfectly well.
You need to move the scanner to the next line. Add this line of code below the error message in the catch section.
input.nextLine();
I want to make it so that a user entering the wrong data type as figureNumber will see a message from me saying "Please enter an integer" instead of the normal error message, and will be given another chance to enter an integer. I started out trying to use try and catch, but I couldn't get it to work.
Sorry if this is a dumb question. It's my second week of an intro to java class.
import java. util.*;
public class Grades {
public static void main(String args []) {
Scanner stdin = new Scanner(System.in);
System.out.println();
System.out.print(" Please enter an integer: ");
int grade = stdin.nextInt();
method2 ();
if (grade % 2 == 0) {
grade -= 1;
}
for(int i = 1; i <=(grade/2); i++) {
method1 ();
method3 ();
}
}
}
public static void main(String args[]) {
Scanner stdin = new Scanner(System.in);
System.out.println();
System.out.print(" Welcome! Please enter the number of figures for your totem pole: ");
while (!stdin.hasNextInt()) {
System.out.print("That's not a number! Please enter a number: ");
stdin.next();
}
int figureNumber = stdin.nextInt();
eagle();
if (figureNumber % 2 == 0) { //determines if input number of figures is even
figureNumber -= 1;
}
for (int i = 1; i <= (figureNumber / 2); i++) {
whale();
human();
}
}
You need to check the input. The hasNextInt() method is true if the input is an integer. So this while loop asks the user to enter a number until the input is a number. Calling next() method is important because it will remove the previous wrong input from the Scanner.
Scanner stdin = new Scanner(System.in);
try {
int figureNumber = stdin.nextInt();
eagle();
if (figureNumber % 2 == 0) { //determines if input number of figures is even
figureNumber -= 1;
}
for(int i = 1; i <=(figureNumber/2); i++) {
whale();
human();
}
}
catch (InputMismatchException e) {
System.out.print("Input must be an integer");
}
You probably want to do something like this. Don't forget to add import java.util.*; at the beginning of .java file.
You want something in the form:
Ask for input
If input incorrect, say so and go to step 1.
A good choice is:
Integer num = null; // define scope outside the loop
System.out.println("Please enter a number:"); // opening output, done once
do {
String str = scanner.nextLine(); // read anything
if (str.matches("[0-9]+")) // if it's all digits
num = Integer.parseInt(str);
else
System.out.println("That is not a number. Please try again:");
} while (num == null);
// if you get to here, num is a number for sure
A do while is a good choice because you always at least one iteration.
It's important to read the whole line as a String. If you try to read an int and one isn't there the call will explode.
You can actually test the value before you assign it. You don't need to do any matching.
...
int figureNumber = -1;
while (figureNumber < 0) {
System.out.print(" Welcome! Please enter the number of figures for your totem pole: ");
if (stdin.hasNextInt()){
figureNumber = stdin.nextInt(); //will loop again if <0
} else {
std.next(); //discard the token
System.out.println("Hey! That wasn't an integer! Try again!");
}
}
...
Hey guys so i've been trying to answer this question for hours:
Write a program that asks the user to input a set of floating-point values. When the
user enters a value that is not a number, give the user a second chance to enter the
value. After two chances, quit reading input. Add all correctly specified values and
print the sum when the user is done entering data. Use exception handling to detect
improper inputs.
I've tried a few different things but i always have the same problem. Once something that isn't a number is given as input, the program outputs the message prompting for another input however the chance is not given, that is to say after 1 incorrect input it prints that message and jumps straight to printing the sum. The best i could do is below i'm just not sure how to approach this problem. Any help is greatly appreciated.
import java.util.Scanner;
import java.util.InputMismatchException;
public class q6{
public static void main(String[] args){
Scanner in = new Scanner(System.in);
boolean firstChance = true;
boolean secondChance = true;
double sum = 0;
while (secondChance){
try{
while (firstChance){
try{
System.out.print("Please enter a number: ");
double input = in.nextDouble();
sum = sum + input;
}
catch (InputMismatchException ex){
firstChance = false;
}
System.out.print("Please enter a number to continue or something else to terminate: ");
double input = in.nextDouble();
sum = sum + input;
firstChance = true;
}
}
catch (InputMismatchException e){
secondChance = false;
}
}
System.out.print("The sum of the entered values is " + sum);
}
}
i'm just not sure how to approach this problem
Pseudocode could be as follows:
BEGIN
MAX_INPUT = 2;
i = 0;
WHILE i < MAX_INPUT
TRY
num = GET_NUM();
CATCH
continue;
FINALLY
i++
END WHILE
END
Since the parsing of input to double is not successful, the scanner does not go past the given input. From javadoc Scanner
Scans the next token of the input as a double. This method will throw
InputMismatchException if the next token cannot be translated into a
valid double value. If the translation is successful, the scanner
advances past the input that matched.
Since the translation is not successful, it does not advance. So, in the catch block you could call in.next() to skip the token.
You could use an integer counter instead of the boolean variables firstChance and secondChance and do something like:
int attempts = 0;
while(attempts < 2) {
try {
//Get user input - possible exception point
//Print sum
} catch(InputMismatchException e) {
attempts++;
//Continue?
}
}
import java.util.*;
public class q6 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
double inputNumber, sum = 0.0;
int correctCount = 0, wrongCount = 0;
while(wrongCount <=1) {
System.out.println("Please enter a numeric floating value:");
try {
inputNumber = in.nextDouble();
correctCount++;
sum += inputNumber;
if(correctCount >= 2)
break;
} catch(InputMismatchException e) {
wrongCount++;
in = new Scanner(System.in);
continue;
}
}
System.out.println(sum);
}
}
You need to break your while loop when you encounter a "bad" input. Then, you'll need to set firstChance to true again, so you can access the second while, you also will need a counter that counts the number of the attempts (I named it chances):
int chances = 0;
while (secondChance){
firstChance = true;
try{
while (firstChance){
try{
System.out.print("Please enter a number: ");
double input = in.nextDouble();
sum = sum + input;
}
catch (InputMismatchException ex){
chances ++;
in = new Scanner(System.in);
firstChance = false;
}
if(!firstChance && chances < 2)
break;
if(chances >= 2) {
System.out.print("Please enter a number to continue or something else to terminate: ");
double input = in.nextDouble();
sum = sum + input;
firstChance = true;
}
}
}
catch (InputMismatchException e){
secondChance = false;
}
}
As stated in my comment, in order to not deal with Scanner wrong read, it would be better to use Scanner#nextLine() and read the data as String, then try to parse it as double using Double#parseDouble(String) since this method already throws NumberFormatException and you can handle this error. Also, it would be better that your code could handle more than 1 request (if your teacher or somebody else asks to do this):
final int REQUEST_TIMES = 2;
double sum = 0;
for(int i = 1; i <= REQUEST_TIMES; i++) {
while (true) {
try {
if (i < REQUEST_TIMES) {
System.out.print("Please enter a number: ");
} else {
System.out.print("Please enter a number to continue or something else to terminate: ");
}
String stringInput = in.nextLine();
double input = Double.parseDouble(stringInput);
sum += input;
} catch (NumberFormatException nfe) {
System.out.println("You haven't entered a valid number.");
break;
}
}
}
The logic is to read the input once, if it is correct then display the sum, else read the input again and display the sum if the input is correct.
public class q6 {
static int trial = 0;
public static void main(String[] args) {
double sum = 0;
while (trial <= 2) {
Scanner in = new Scanner(System.in);
try {
trial++;
System.out.print("Please enter a number: ");
double input = in.nextDouble();
sum = sum + input;
trial=3;
} catch (InputMismatchException ex) {
trial++;
if (trial == 4){
System.out.println("You have entered wrong values twice");
}
}
}
if (trial <= 3){
System.out.print("The sum of the entered values is " + sum);
}
}
The above program is just a rough idea, you can improve this to adapt to your need.
I have been trying to stop the exceptions but I cannot figure out how.
I tried parseInt, java.util.NormalExceptionMismatch etc.
Does anyone have any insight how to fix this problem? Formatting is a bit off due to copy and paste.
do
{
System.out.print(
"How many integers shall we compare? (Enter a positive integer):");
select = intFind.nextInt();
if (!intFind.hasNextInt())
intFind.next();
{
// Display the following text in the event of an invalid input
System.out.println("Invalid input!");
}
}while(select < 0)
Other methods I have tried :
do
{
System.out.print(
"How many integers shall we compare? (Enter a positive integer):");
select = intFind.nextInt();
{
try{
select = intFind.nextInt();
}catch (java.util.InputMismatchException e)
{
// Display the following text in the event of an invalid input
System.out.println("Invalid input!");
return;
}
}
}while(select < 0)
It seems to me that you want to skip everything until you get an integer. This code here skips any input except an integer.
As long as there is no integer available (while (!in.hasNextInt())) discard the available input (in.next). When integer is available - read it (int num = in.nextInt();)
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (!in.hasNextInt()) {
in.next();
}
int num = in.nextInt();
System.out.println("Thank you for choosing " + num + " today.");
}
}
Quick sample of how to catch exceptions:
int exceptionSample()
{
int num = 0;
boolean done = false;
while(!done)
{
// prompt for input
// inputStr = read input
try {
num = Integer.parseInt(inputStr);
done = true;
}
catch(NumberFormatException ex) {
// Error msg
}
}
return num;
}
IMO, the best practice is to use nextLine() to get a String input, then parseInt it to get the integer. If unparsable, just complain back to the user and request re-entry.
Remember you may have to do a second nextLine() (discard the input) to clear up the buffer.