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Try Catch InputMismatchException not actually catching my Error

I have copied the exact same code from another program. Whenever inputting deliberate incorrect results the InputMismatchException still occurs and the program crashes.
import java.util.*;
public class Runner{
public static void main (String args[]){
Scanner sc = new Scanner(System.in);
sc.useDelimiter("\n");
Fixture f = new Fixture();
boolean inputValid = false;
int choice = 0;
do{
do {
System.out.println("\f\t\tFootball Database");
System.out.println("A utility to help make footballing events easier to manage.");
System.out.println("");
System.out.println("\t> Press 1 to manage players ");
System.out.println("\t> Press 2 to manage teams");
System.out.println("\t> Press 3 to manage coaches");
System.out.println("\t> Press 4 to manage fixtures");
System.out.println("\t> Press 5 to save database to file");
System.out.println("\t> Press 6 to load database from file");
System.out.println("\t> Press 7 to terminate program");
System.out.println("");
System.out.println("");
System.out.println("©Thomas Camilleri 2017");
try{
choice = sc.nextInt();
inputValid = true;
}catch(InputMismatchException e){
System.out.println("Invalid input");
inputValid = false;
sc.nextInt();
sc.nextInt();
}
}while(inputValid == false);

Here is the salient part of your code:
try {
choice = sc.nextInt(); // NOT HERE
inputValid = true;
} catch(InputMismatchException e){
System.out.println("Invalid input");
inputValid = false;
sc.nextInt(); // HERE
sc.nextInt();
}
If you look at the stacktrace that you got, and look at the line numbers, you will see that the line in your code where the exception happens is the one I have tagged with // HERE.
(Compile and run the original program and look at the stacktrace to see what I mean. Compare the line numbers in the stack trace with the source code.)
As you can see, that line is NOT in the try { ... } block. It is in the exception handler block.
What has happened is that you have caught the exception that was thrown at the line tagged // NOT HERE, and then you have called sc.nextInt() again (at // HERE). The second call has simply attempted to read the same input characters again.
The behavior of the nextInt method is as follows:
get characters sufficient to form a token
attempt to convert the entire token into an integer (using base-10 integer syntax)
if the conversion succeeds, return the converted integer
if the conversion fails, put all of the characters back and then throw an exception.
I strongly encourage you to carefully read the javadocs for the Scanner class so that you understand what the methods you are using actually do.
So ... as you see ... if you just call nextInt() after a failed nextInt() call, you just get the same failure repeated.
The reason that second exception is not caught is that it has not been thrown within the try { ... } block.
Solution: Instead of calling nextInt() in the handler, you should call a method that is going to just discard the rubbish. In this case, the most sensible thing to do is to throw away everything up to the next end-of-line.
Hint #1: the nextLine() gets everything up to the next end-of-line. Read the javadocs for that method too.
Hint #2: if you understand what I said, you will know where to put your modification.

Try like:
choice = Integer.parseInt(sc.nextLine());
and your program will like this:
try (Scanner sc = new Scanner(System.in)) {
sc.useDelimiter("\n");
// Fixture f = new Fixture();
boolean inputValid = false;
int choice = 0;
// removed outer do..while(); loop
do {
System.out.println("\f\t\tFootball Database");
System.out.println("A utility to help make footballing events easier to manage.");
System.out.println("");
System.out.println("\t> Press 1 to manage players ");
System.out.println("\t> Press 2 to manage teams");
System.out.println("\t> Press 3 to manage coaches");
System.out.println("\t> Press 4 to manage fixtures");
System.out.println("\t> Press 5 to save database to file");
System.out.println("\t> Press 6 to load database from file");
System.out.println("\t> Press 7 to terminate program");
System.out.println("");
System.out.print("Enter your choice : ");
try{
// Always use nextLine() if you mix String and basic Datatype
choice = Integer.parseInt(sc.nextLine());
inputValid = true;
}catch(NumberFormatException e){
System.out.println("Invalid input");
inputValid = false;
// Removed unnecessary two sc.nextInput() lines
}
}while(inputValid == false);
System.out.println("choice is : " + choice);
}

User input and Exception in Java

I want to add an integer to a list based on user input. The user has to type all the integers he/she wishes then press enter. if they finish inputting integer, they are supposed to press the "enter" button without typing anything.
I have made my code, but there are several mistakes
the exception keeps popping up because every time say for example I enter integer 10, then I finish. I press "enter" with nothing. this raises the exception. how do I tackle this problem?
and another thing, how do I make the program so that if the user puts invalid input, instead of crashing or breaking. It asks the user again to prompt the correct input.
this is what I have done
package basic.functions;
import java.util.*;
import java.text.DecimalFormat;
public class Percent {
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
reader.useDelimiter(System.getProperty("line.separator"));
List<Integer> list = new ArrayList<>();
System.out.println("Enter Integer: ");
while (true) {
try {
int n = reader.nextInt();
list.add(Integer.valueOf(n));
} catch (InputMismatchException exception) {
System.out.println("Not an integer, please try again");
break;
}
}
reader.close();
}
}
output
Enter Integer:
10
Not an integer, please try again
[10]

I'd suggest you utilise Scanner#hasNextInt to identify whether an integer has been entered or not. As for when the "user presses enter without typing anything", we can simply use the String#isEmpty method.
while (true) {
if(reader.hasNextInt()) list.add(reader.nextInt());
else if(reader.hasNext() && reader.next().isEmpty()) break;
else System.out.println("please enter an integer value");
}
note - in this case, you don't need to catch InputMismatchException because it won't be thrown.

while (true) is generally a bad sign, if you ever have that in your code you are almost certainly wrong.
What you probably want is something like this:
String input;
do {
input = reader.next();
// Parse the input to an integer using Integer.valueOf()
// Add it to the list if it succeeds
// You will need your try/catch etc here
while (!input.isEmpty());
Here the loop is checking the exit condition and running until it meets it. Your processing is still done inside the loop as normal but the program flow is a lot cleaner.

How to ask te user and depending on his response run all the code again?

I'm a newbie in Java. I started these days and I'm practicing the catch and try exception. I have this code below which solve an operation between to numbers and I'd like to know what can I do, if for example I want that the user, once he makes an operation and get his result, that this has the possibility to make another operation. something like comes up a question asking if he wants to realize another problem and the code run again from the beginning.
package justpractice;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner Operation = new Scanner(System.in);
int x=1;
while(x==1){
try{
System.out.println("Insert numerator");
int n1 = Operation.nextInt();
System.out.println("Insert denominator");
int n2=Operation.nextInt();
double division = n1/n2;
System.out.println(division);
x=2;
}
catch(Exception e){
System.out.println("Insert a valid value");
}
}
}
}

You can do, for example, adding an if statement with a
System.out.println("Do you want to recalculate ? (1/0 Yes/No)");
Operation.nextInt();
then if the input is 1, keep x = 1, else do x = 2.

Try this code amendment;
package justpractice;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner Operation = new Scanner(System.in);
//better practice
while(true){
try{
System.out.println("Insert numerator");
int n1 = Operation.nextInt();
System.out.println("Insert denominator");
int n2=Operation.nextInt();
double division = n1/n2;
System.out.println(division);
System.out.println("Continue? (y/n)");
String response = Operation.nextLine();
if (response.equals("n")){
break;
}
}
catch(Exception e){
System.out.println("Insert a valid value");
}
}
}
}

To allow your user to calculate division again, you could use a do-while loop. With this loop you can execute the code inside once, ask the user if they would like to calculate again, and repeat the code if they do.
An outline of your code with the loop would something like this:
...
boolean again = true;
do { //Main loop
try {
... //Your division code
... //Put code to ask the user if they want to calculate again here
} catch(Exception e) {
...
}
} while(again == true); //Repeat if the user wants to continue
To get input on if the user wants to calculate again, I recommend using another do-while loop with your Scanner. This loop will allow you to repeat the code when the answer is invalid. In this case, when it's not "y" or "n".
String input;
do {
System.out.println("Would you like to continue? (y/n)");
input = operation.next();
} while(!input.equalsIgnoreCase("y") && !input.equalsIgnoreCase("n"));
After you have got the user's input, you still need to terminate the loop if they said "n". To do this you could use an if statement.
if(input.equalsIgnoreCase("n")) {
again = false; //Terminate the main loop
operation.close(); //Prevent any resource leaks with your Scanner
}
There is no need to check if the user input "y" as again is set to true by default.
Side Note: Variables should always be camelCase. Look at the Java Naming Conventions to learn more about naming things in Java.
EDIT:
The reason the console is repeatedly logging that you entered a non-number even though you entered it once, I'm not exactly sure. I think it's because the call to nextInt() never finishes because of the InputMismatchException being thrown, causing the next call to nextInt() (After the do-while repeats) to think that the letter/symbol you just entered is the one you want to process, repeating the exception over and over again.
To solve this, add this line into your catch block:
if(operation.hasNext()) operation.next();
This will call next() and complete the process of marking the letter/symbol you just entered as already processed, then repeat the do-while loop as normal.

Switch returning same statement multiple times

I'm making a school assignment and this time around I thought about using a switch statement since it looked more efficient.
It's just something basic but if I enter a letter for example and after that number 1 for example it would return case 1 twice?
This is my code for the entire class so far:
import java.util.InputMismatchException;
import java.util.Scanner;
public class Test {
private int option;
public static void main(String[] args) {
Test t = new Test();
t.start();
t.optionMenu();
}
public void start() {
System.out.println("Make your choice:");
System.out.println("1: Play");
System.out.println("2: Options");
System.out.println("3: Exit");
}
public void optionMenu() {
try {
Scanner sc = new Scanner(System.in);
this.option = sc.nextInt();
System.out.println(this.option);
} catch (InputMismatchException e) {
System.out.println("Please enter a number");
optionMenu();
}
switch (this.option) {
case 1:
System.out.println("Game starting...");
break;
case 2:
System.out.println("Loading options");
break;
case 3:
System.out.println("Game exiting...");
System.exit(0);
break;
default:
System.out.println("Enter a valid number (1, 2 or 3");
break;
}
}
}
Any help would be much appreciated, thanks!

When you call sc.nextInt() without first asking if (sc.hasNextInt()), you are open to some strange behavior when end-users start typing unexpected input, such as letters. In this case the scanner would not advance its reading pointer, so your program will get stuck reading the same incorrect output.
To fix this issue, add a loop that "clears out" the invalid entry before attempting to read an int again, like this:
while (!sc.hasNextInt()) {
System.out.print("You need to enter an integer.");
sc.nextLine(); // Clear out the bad input
}
int val = sc.nextInt(); // At this point we know that sc.hasNextInt(), because that's the loop condition
Another point is that it is not a good idea to do with recursion what can be done with iteration: the recursive call to optionsMenu is going to accumulate as many levels of invocation as the number of times the end-user enters an incorrect value, so a very persistent user could theoretically force a stack overflow on your program by entering invalid data repeatedly.
Using the code fragment above would free you from the need to call optionsMenu recursively, and also from catching the input exception.

It's just something basic but if I enter a letter for example and after that number 1 for example it would return case 1 twice?
I'm not sure what you mean here. Firstly, your idea works, this code should be fine!
Second, if you enter anything besides just the number 1, 2, or 3, you will go to the "default:" block of code. Since you are prompting the user again if they fail, typing "a" or "a1" into the prompt just shows the menu again. The user needs to just type "1", "2", or "3" to successfully select a menu option.

Exception Handling with wrong User Input Java

I need help Using Exception handling with Wrong User Input. I am creating a text based game that welcomes the User and then goes to the main menu. It then tell the User the options, and then look for User input. For some Reason, whenever I input 1 or 2, It says: "Your input is invalid, please try again" And goes back to the choices. I don't know exactly where I am going wrong, hopefully someone can help me. Also, it won't catch the Mismatch Exception either. Hope you can Help! Thanks, Shandan
public static void main(String[] args) {
System.out.println("Welcome to Spec Ops!");
System.out.println("Please state your name:");
Scanner name = new Scanner(System.in);
String Name = name.next();
System.out.println("Hello "+Name);
mainMenu();
}
public static void mainMenu() {
System.out.println("1. Story Mode");
System.out.println("2. Infinant Combat");
Scanner input = new Scanner(System.in);
Object Selection = input.nextInt();
boolean validOption = true;
Integer x;
try {
x = (Integer)Selection;
} catch(ClassCastException cce){
System.out.println("Your input is invalid, please try again");
validOption = false;
} catch(InputMismatchException ime){
System.out.println("Your input is invalid, please try again");
validOption = false;
}
if(validOption) {
System.out.println("Hello!");
}
else {
mainMenu();
}
}
}

Scanner.nextInt returns ant int, so there is no need to go Object selection = scanner.nextInt(), and then cast to an int, you can merely have int selection = scanner.nextInt() and surround that in a try catch that chatches java.util.InputMismatchException, which is the exception thrown when the user types a letter and not an number

You can temporarily change your code and use the Pokemon exception handling and check the stack trace to see what kind of exception you should implement:
try {
//do something
} catch(Exception e) {
//Gotta catch 'em all!
e.printStackTrace();
}
Once you know the exact exception, put some breakpoints and refactor your code accordingly.
Also, you don't have to control the flow of your program by changing your boolean variable inside the catch blocks:
boolean isValidOption = false;
Integer x;
try {
x = (Integer)Selection;
isValidOption = true;
} catch...

When you read the name in you are using scanner.next(). This does not read in the end of line character, so when you call scanner.nextInt() it reads the new line and fails to parse as an int.
You should change the name reading to scanner.nextLine() and if you want to handle invalid int inputs from the user, the call to nextInt needs to be in your try block and catch the InputMismatchException instead of the ClassCastException.