I am trying to create a code where an int is supposed to be entered and then have exceptions for if the int is not between 9 and 99, another exception if a double is entered instead of int and then a third exception if a string is entered. How do i do this? i have below what i have so far but am not sure how to correct it. thanks
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean correct = true;
do {
try {
System.out.println("Enter an Integer between 9 and 99");
int number = input.nextInt();
if (number >= 9 && number <= 99) {
System.out.println("Thank you, Initialization completed");
correct = false;
} else if (number < 9 || number > 99) {
throw new Exception("Integer is not within the range");
}
if (input.hasNextDouble()) {
throw new Exception("Integer not entered");
} else {
correct = false;
}
if (input.hasNext("")) {
throw new NumberFormatException("Integer not entered");
} else {
correct = false;
}
} // check for range
catch (Exception e1) {
System.out.println("Number is not within 9 and 99");
System.out.println();
input.nextLine();
} catch (Exception e2) {
System.out.println("An integer was not entered");
System.out.println();
input.nextLine();
} catch (NumberFormatException e3) {
System.out.println("An integer was not entered");
System.out.println();
input.nextLine();
}
} while (correct);
}
Method .getMessage() returns the string given in constructor:
throw new Exception("HERE");
When you catch Exception, you catch also NumberFormatException, InputMismatchException, etc.
so you must catch broader ones last.
catch (NumberFormatException e3) { // Precisier goes first
System.out.println("An integer was not entered");
System.out.println();
input.nextLine();
}
catch (Exception e1) {
System.out.println(e1.getMessage());
System.out.println();
input.nextLine();
}
Related
This question already has answers here:
What is a NumberFormatException and how can I fix it?
(9 answers)
Closed 6 years ago.
I'm not sure how to set it up, but my method has a user input a number between a minimum and maximum, although I have no idea how to handle the NumberFormatException with a try-catch block. Are there any suggestions in regards to how this can be fixed?
static int promptForInt(String prompt, int min, int max){
System.out.println(prompt);
String input = in.readLine();
int parsedInt = Integer.parseInt(input);
while(!(parsedInt > min && parsedInt < max)){
System.out.println("Your input is invalid. " + prompt);
input = in.readLine();
parsedInt = Integer.parseInt(input);
}
return parsedInt;
}
static int promptForInt(String prompt, int min, int max){
System.out.println(prompt);
String input = in.readLine();
int parsedInt;
boolean exceptionThrown = false;
do {
try {
parsedInt = Integer.parseInt(input);
} catch(NumberFormatException e) {
exceptionThrown = true;
}
if (exceptionThrown || (!(parsedInt > min && parsedInt < max)) {
System.out.println("Your input is invalid. " + prompt);
input = in.readLine();
parsedInt = Integer.parseInt(input);
} else {
return parsedInt;
}
} while(true)
}
From my post about NumberFormatException:
Ad. 4.
Finally we come to the place in which we agree, that we can't avoid situations when it's user typing "abc" as a numeric string. Why? Because he can. In a lucky case, it's because he's a tester or simply a geek. In a bad case it's the attacker.
What can I do now? Well, Java gives us try-catch you can do the following:
try {
i = Integer.parseInt(myString);
} catch (NumberFormatException e) {
e.printStackTrace();
//somehow workout the issue with an improper input. It's up to your business logic.
}
Exceptions
In Java Exceptions are used for marking unexpected situations. For example parsing non-numeric String to a number (NumberFormatException) or calling a method on a null reference (NullPointerException). You can catch them in many ways.
try{
//some code
} catch (NumberFormatException e1) {
e.printStackTrace() //very important - handles the Exception but prints the information!
} catch (NullPointerException e2) {
e.printStackTrace();
}
or using the fact, that they all extend Exception:
try {
//somecode
} catch (Exception e) {
e.printStackTrace;
};
or since Java 7:
try {
//somecode
} catch (NullPointerException | NumberFormatException e) {
e.printStackTrace;
};
Exceptions can be easier to deal with if you think of them as the method "Trying to do something, but it couldn't because of X". X is the exception.
The below code could be one way you modify your code to handle the exception:
static int promptForInt(String prompt, int min, int max) {
Integer parsedInt = null; // use an Integer so null can be used to mean an invalid value
while (parsedInt == null) {
System.out.println(prompt);
String input = in.readLine();
int temp;
try {
temp = Integer.parseInt(input);
} catch(NumberFormatException e) {
System.out.print(input+" is not a number. ");
continue;
}
if (temp < min || temp > max) {
System.out.print("Your number must be between "+min+" and "+max+" (inclusive). ");
} else {
parsedInt = temp;
}
}
return parsedInt;
}
Some things you should notice: Firstly, you have not defined in. You could do that like so:
BufferedReader in = new BufferedReader(new InputStreamReader(System.in, "UTF-8"));
If you do that, you'll soon see that you have another Exception to deal with: this can throw UnsupportedEncodingException, and also readLine can throw IOException.
Your method must return a valid integer or it will not exit (you really should supply the user some means of exiting the loop without entering a valid number). Since that's not going to be possible if, for instance, you couldn't read anything from System.in your method needs a reasonable way of telling the caller: "I tried to get an int from the user, except I was stopped by X".
You may actually end up doing something more like:
static int promptForInt(String prompt, int min, int max) throws UserInputException {
BufferedReader in;
try {
in = new BufferedReader(new InputStreamReader(System.in, "UTF-8"));
} catch(UnsupportedEncodingException unsupported) {
// create a new exception and supply the cause as an inner exception
throw new UserInputException("Could not open a reader for System.in", unsupported);
}
Integer parsedInt = null; // use an Integer so null can be used to mean an invalid value
while (parsedInt == null) {
System.out.println(prompt);
String input;
try {
input = in.readLine();
} catch (IOException ioException) {
throw new UserInputException("Could not read a line", ioException);
}
if (input.length() == 0) {
throw new UserInputException("User aborted input");
}
int temp;
try {
temp = Integer.parseInt(input);
} catch(NumberFormatException e) {
System.out.print(input+" is not a number. ");
continue;
}
if (temp < min || temp > max) {
System.out.print("Your number must be between "+min+" and "+max+" (inclusive). ");
} else {
parsedInt = temp;
}
}
return parsedInt;
}
How would I use InputMismatchException to determine if the value entered into the Scanner is not an integer? Basically, if they enter in a word instead of an integer I need to use InputMismatchException to return a message.
while (true) {
Scanner sc = new Scanner(System.in);
int i = sc.nextInt();
try{
Integer.parseInt(i);
} catch (InputMismatchException e) {
System.out.println("Sorry, " + i + " is not a number.");
}
if (i == 1) {
System.out.println("1 was selected");
} else {
System.out.println("1 was not selected");
}
Change your code as such:
while (true) {
Scanner sc = new Scanner(System.in);
String s = sc.nextLine();
try{
int i = Integer.parseInt(s);
if (i == 1) {
System.out.println("1 was selected");
} else {
System.out.println("1 was not selected");
}
} catch (NumberFormatException e) {
System.out.println("Sorry, " + s + " is not a number.");
}
}
Changes:
Note use of nextLine(). This is because it seems you want to use the input as part of your error message, and nextInt() won't let you do that.
We can now move the i declaration inside the try block, along with
the code that uses it, so we won't issue "1 was not selected" when the input was "ssss" or whatever.
We use NumberFormatException, as that's what Integer.parseInt()
throws when it can't parse an integer from the String.
This is what i mean
while (true) {
Scanner sc = new Scanner(System.in);
int i = -1;
try
{
i = sc.nextInt();
}
catch (InputMismatchException e)
{System.out.println("Sorry, " + i + " is not a number.");}
if (i == 1)
System.out.println("1 was selected");
else
System.out.println("1 was not selected");
}
I am new to the programming, I think I got the code for your question.
while (true) {
Scanner sc = new Scanner(System.in);
String s = sc.nextLine();
try{
int i = Integer.parseInt(s);
if (i == parseInt(i, 3)) {
System.out.println(s+" is selected");
} else {
System.out.println("Input value is " + s);
}
} catch (NumberFormatException e) {
System.out.println("Sorry, " + s + " is not a number.");
}}}
private static int parseInt(int i, int j) {
// TODO Auto-generated method stub
return 0;
}}
reference
import java.util.Scanner;
public class Lab4_5 {
public static void main(String[]args) {
Scanner scan= new Scanner(System.in);
int rows=0;
int rowIndex=0, colIndex=0;
boolean choice1= true;
String y="y";
String n="n";
boolean first = true;
while (choice1==true) {
if (first==true) {
first=false;
System.out.println("Do you want to start(Y/N): ");
} else if (first==false) {
System.out.println("Do you want to continue(Y/N): ");
}
String choice2=scan.next();
if (choice2.equals(y)) {
System.out.println("How many rows/columns(5-21)?");
rows=scan.nextInt();
while (rows<5 || rows>21) {
System.out.println("That is either out of range or not an integer, try again! ");
rows=scan.nextInt();
}
System.out.println("What character?");
String choice3=scan.next();
System.out.println(" ");
for (rowIndex=1; rowIndex<=rows; rowIndex++) {
for (colIndex=1; colIndex<=rows; colIndex++) {
if (rowIndex==1 || rowIndex==rows || colIndex==1 || colIndex==rows) {
System.out.print(choice3);
} else {
System.out.print(" ");
}
}
System.out.println();
}
} else if(choice2.equals(n)) {
choice1 = false;
System.out.println("Thank you. Goodbye.");
} else {
System.out.println("Please either enter Y or N.");
}
}
}//end of main
}
The code prints what I need it to print, but I also have to have something in the code when it asks how many rows/columns to catch whether or not i input something other than an integer(in the part below). need some help, we haven't done anything yet with how to catch exceptions and i don't know how to start.
String choice2=scan.next();
if (choice2.equals(y)) {
System.out.println("How many rows/columns(5-21)?");
rows=scan.nextInt();
while (rows<5 || rows>21) {
System.out.println("That is either out of range or not an integer, try again! ");
rows=scan.nextInt();
}
}
You need to understand this please look into it.
Basic understanding is
try {
//Something that can throw an exception.
} catch (Exception e) {
// To do whatever when the exception is caught.
}
There is also an finally block which will always be execute even if there is an error. it is used like this
try {
//Something that can throw an exception.
} catch (Exception e) {
// To do whatever when the exception is caught & the returned.
} finally {
// This will always execute if there is an exception or no exception.
}
In your particular case you can have the following exceptions (link).
InputMismatchException - if the next token does not match the Integer regular expression, or is out of range
NoSuchElementException - if input is exhausted
IllegalStateException - if this scanner is closed
So you would need to catch exceptions like
try {
rows=scan.nextInt();
} catch (InputMismatchException e) {
// When the InputMismatchException is caught.
System.out.println("The next token does not match the Integer regular expression, or is out of range");
} catch (NoSuchElementException e) {
// When the NoSuchElementException is caught.
System.out.println("Input is exhausted");
} catch (IllegalStateException e) {
// When the IllegalStateException is caught.
System.out.println("Scanner is close");
}
You can create a try-catch block like so:
try {
int num = scan.nextInt();
} catch (InputMismatchException ex) {
// Exception handling here
}
If you want to implement this in your code, I suggest doing this:
while (true) {
try {
rows = scan.nextInt();
if (rows<5||rows>21) {
break;
}
else {
System.out.println("That is either out of range or not an integer, try again! ");
}
} catch (InputMismatchException ex) {
System.out.println("That is either out of range or not an integer, try again! ");
}
}
See here for more details.
String choice2=scan.next();
if(choice2.equals(y)){
System.out.println("How many rows/columns(5-21)?");
try
{
rows=scan.nextInt();
}catch(Exception e)
{
rows = -1;
}
while(rows<5||rows>21){
System.out.println("That is either out of range or not an integer, try again! ");
try
{
rows=scan.nextInt();
}catch(Exception e)
{
rows = -1;
}
}
I would like to know how to set up a try/catch on a variable that is a string but has to be checked against an integer like so:
public static void isValidAge(String age) {
int age2 = Integer.parseInt(age);
try {
if(age2 <= 0 || age2 >= 150) {
throw new NumberFormatException();
}
}
catch (NumberFormatException ex) {
if(age2 <= 0) {
System.err.print("Age can not be 0 or negative.");
}
else if (age2 >= 150) {
System.err.print("Age can not be equal to or more than 150.");
}
else if (age.contains("##$")) {
System.err.print("You did not enter a valid age.");
}
}
}
The try/catch must also be able to handle characters and keep the program running still.
The try-catch should be on the parse attempt itself:
int age2 = -1; //set to an invalid value
try
{
age2 = Integer.parseInt(age);
}
catch(Exception ex)
{
System.out.println("Error: Could not parse age to number, exiting");
return; //exit function
}
So I'm making this school assignment and basically need to make sure that the user inputs a valid option (1, 2 or 3).
I should of used switches but this is what I made:
private void choice() {
try {
Scanner s = new Scanner(System.in);
int option = s.nextInt();
if (option == 1) {
start();
}
if (option == 2) {
info();
}
if (option == 3) {
System.exit(0);
}
throw new InputMismatchException("Enter valid input");
} catch (InputMismatchException e) {
System.out.println(e.getMessage());
}
}
If you enter a number in the console it will return my message, if you enter anything else it will return "null". Why is that? Because if I remove the exception and look at the stacktrace (when entering a letter for example) it shows an InputMismatchException.
Thanks in advance!
public void choice() throws InputMismatchException {
Scanner s = new Scanner(System.in);
int option = 0;
try {
option = s.nextInt();
} catch (InputMismatchException e) {
throw new InputMismatchException("Enter valid input");
}
if (option == 1) {
System.out.println("e");
} else if (option == 2) {
System.out.println("f");
} else if (option == 3) {
System.exit(0);
} else {
throw new InputMismatchException("Enter valid input");
}
}
public static void main(String[] args) {
try {
new Test().choice();
} catch (InputMismatchException e) {
System.out.println(e.getMessage());
}
}
Put int option, inside the try. Done.