Easier way to guarantee integer input through Scanner? - java

For a program I am writing, I need to ask a user for an integer between 1 and 8. I've tried multiple (cleaner) ways of doing this but none of them worked, so I'm left with this:
int x = 0;
while (x < 1 || x > 8)
{
System.out.print("Please enter integer (1-8): ");
try
{
x = Integer.parseInt(inputScanner.next());
}
catch(NumberFormatException e)
{
x = 0;
}
}
Where inputScanner is a Scanner. Surely there is a better way?

Scanner does regular expressions, right? Why not check if it matches "^[1-8]$" first?

Using the nextInt() is already an improvement compare to simply using the next() method. And before that, you can use the hasNextInt() to avoid haing all this bunch of useless exceptions.
Resulting in something like this:
int x = 0;
do {
System.out.print("Please...");
if(scanner.hasNextInt()) x = scanner.nextInt();
else scanner.next();
} while (x < 1 || x > 8);

I had to do a graphic interface calculator (works only with Integers), and the problem was, that
the Tests didn't allow any Exceptions to be thrown if the input wasn't
Integer. So I couldn't use
try { int x = Integer.parseInt(input)} catch (Exception e) {dosomethingelse}
Because Java programs generally treat an input to a JTextField as a String
I used this:
if (input.matches("[1-9][0-9]*"){ // String.matches() returns boolean
goodforyou
} else {
dosomethingelse
}
// this checks if the input's (type String) character sequence matches
// the given parameter. The [1-9] means that the first char is a Digit
// between 1 and 9 (because the input should be an Integer can't be 0)
// the * after [0-9] means that after the first char there can be 0 - infinity
// characters between digits 0-9
hope this helps :)

You could try something like this:
Scanner cin = new Scanner(System.in);
int s = 0;
boolean v = false;
while(!v){
System.out.print("Input an integer >= 1: ");
try {
s = cin.nextInt();
if(s >= 1) v = true;
else System.out.println("Please input an integer value >= 1.");
}
catch(InputMismatchException e) {
System.out.println("Caught: InputMismatchException -- Please input an integer value >= 1. ");
cin.next();
}
}

Apache Commons is your friend. See NumberUtils.toInt(String, int)

String input;
int number;
while (inputScanner.hasNextLine())
{
input = inputScanner.nextLine();
if (input.equals("quit")) { System.exit(0); }
else
{
//If you don't want to put your code in here, make an event handler
//that gets called from this spot with the input passed in
try
{
number = Integer.parseInt(input);
if ((number < 1) || (number > 8))
{ System.out.print("Please choose 1-8: "); }
else { /* Do stuff */ }
}
catch (NumberFormatException e) { number = 0; }
}
}
I always like to pull in the full string so you can be sure that the user pushed the Enter button. If you just use inputScanner.nextInt() you can put two ints on a line and it will pull in one, then the other.

Example code:
int x;
Scanner in = new Scanner(System.in);
System.out.println("Enter integer value: ");
x = in.nextInt();
An array can also be used to store the integer.

Related

Asks user to enter values until an integer is entered

I am a noob in programming.
I wanted to write code for a prog which asks user to enter value until an integer is entered.
public class JavaApplication34 {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int flag = 0;
while(flag == 0) {
int x = 0;
System.out.println("Enter an integer");
try {
x = sc.nextInt();
flag = 1;
} catch(Exception e) {
System.out.println("error");
}
System.out.println("Value "+ x);
}
}
}
I think the code is correct and it should ask me to enter the value again if i have entered anything other than an integer.
But when i run it , and say i enter xyz
it iterates infinite time without asking me to enter the value.
test run :
Enter an integer
xyz
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
Enter an integer
error
Value 0
When a scanner throws an InputMismatchException, the scanner will not
pass the token that caused the exception.
Hence sc.nextInt() reads the same token again and throws the same exception again.
...
...
...
catch(Exception e){
System.out.println("error");
sc.next(); // <---- insert this to consume the invalid token
}
You can change your logic as shown below :
int flag = 0;
int x = 0;
String str="";
while (flag == 0) {
System.out.println("Enter an integer");
try {
str = sc.next();
x = Integer.parseInt(str);
flag = 1;
} catch (Exception e) {
System.out.println("Value " + str);
}
}
Here we have first read the input from Scanner and then we are trying to parse it as int, if the input is not an integer value then it will throw exception. In case of exception we are printing what user has enter. When user enters an integer then it will parsed successfully and value of flag will update to 1 and it will cause loop to exit.
In the error case, you need to clear out the string you've entered (for instance, via nextLine). Since it couldn't be returned by nextInt, it's still pending in the scanner. You also want to move your line outputting the value into the try, since you don't want to do it when you have an error.
Something along these lines:
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int flag = 0;
while(flag == 0)
{
int x = 0;
System.out.println("Enter an integer");
try
{
x = sc.nextInt();
flag = 1;
System.out.println("Value "+ x);
}
catch (Exception e){
System.out.println("error");
if (sc.hasNextLine()) { // Probably unnecessary
sc.nextLine();
}
}
}
}
Side note: Java has boolean, there's no need to use int for flags. So:
boolean flag = false;
and
while (!flag) {
and
flag = true; // When you get a value
The answers to this question might help you
It makes use of Scanners .hasNextInt() function!

validate data type at the time of user input

I want to validate the data type, (in my case it is 'int'), at the time of user input using Scanner.
I wrote code below.
public static void main(String[] args) throws IOException {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter size of an array...");
int n = 0;
// 1st block, n <= 100
do {
System.out.println("Capacity must be less than or equal to 100");
try {
n = scanner.nextInt();
} catch (InputMismatchException e) {
System.out.println("Enter only integers ");
}
} while (n > 100);
int[] arr = new int[n];
// block 2. Value must be greater than 0 and less than 10000
for (int i = 0; i < n;) {
do {
try {
arr[i] = scanner.nextInt();
} catch (InputMismatchException e) {
System.out.println("Enter only integer value");
}
} while (arr[i] > 10000 || arr[i] < 1);
i++;
}
scanner.close();
for (int i = n - 1; i >= 0; i--)
System.out.println(arr[i]);
}
}
Issue is,
in 1st block, if i give character, program terminates. "How to keep loop running on failed validation??"
in 2nd block if i give non integer, it runs infinitely with message, "Enter only integer value".
From debug, i conclude that, Without waiting for input it takes last non-int value which was provided before.
Why compiler is taking last value??
Any suggestion ?
1) You assign 0 as default value of the n integer you are using to get the user input : int n = 0;
So if the input triggers a InputMismatchException, you arrive in the while statement with a n that equals 0 and while (n > 100) with n = 0 is false.
So you exit the loop.
To solve this problem :
use a wrapper Integer : Integer n = null; that has a nullable value allows to know if no value was accepted in the scanner reading
change the while statement condition to check your requirement :
while (n == null || n > 100);
2) For first case as for the second case, if the input doesn't match with the type required (here a int value), the current token in the scanner is not read.
Just ignore this token to avoid entering in an infinite loop :
catch (InputMismatchException e) {
...
scanner.nextLine();
}
As an addition to davidxxx's answer:
Here's the same code using Scanner.hasNextInt() method.
Scanner scanner = new Scanner(System.in);
System.out.println("Enter size of an array...");
while (!scanner.hasNextInt()) scanner.next();
int arrayLength = scanner.nextInt();
int[] arr = new int[arrayLength];
boolean arrayFull = false;
int index = 0;
System.out.println("Enter array values");
while (!arrayFull){
while (!scanner.hasNextInt()) scanner.next();
int value = scanner.nextInt();
if (value < 0 || value > 1000) {
System.out.println("Enter only integer value < 0 and > 1000");
continue;
}
arr[index++] = value;
if (index >= arrayLength){
arrayFull = true;
}
}

Using java scanner to check two conditions while taking user input

I need to user to enter an int between 1 and 301.
I have this simple loop here to check for user input.
I just want a single number from the user, and if the user enters anything other than an int between 1 and 301, I want to display the print line and prompt the users to try again until they enter a valid input.
while (!sc.hasNextInt()) {
System.out.print("Invalid Input. Please enter a valid number between 1 and 301: ");
sc.next();
}
int numToCheck = sc.nextInt();
// do stuff with numToCheck
This checks that the input is an int, but I can't seem to find a way to give the int input a bound. I tried to assign the user input to a variable and then check the conditions input < 1 or input > 301, but I get InputMismatchException if user enters a letter. How should I store the user input? (I want to store it as an int to check the conditions, but can't do that since I don't know what the user will enter).
Perhaps there is a better design to accomplish all this. Those are welcomed too.
Thanks in advance.
You're not saving the value of the of the input. So your program is waiting on the user to enter a number each time it see "sc.nextInt()" Assign the input to a variable, and then check the condition.
EDIT: okay, I'll go the extra mile for you. See if this works.
***Accounted for the case where the user might enter a character instead of a number.
import java.util.*;
public class HelloWorld{
public static void main(String []args){
Scanner sc = new Scanner(System.in);
int input;
while (true){
if (sc.hasNextInt()){
input = sc.nextInt(); // Assign the next integer to a variable
if (input <= 301 && input >= 1){ // Check if integer meets condition
break; // Condition met, break out of loop
}
}else{
sc.next();
}
System.out.println("Invalid Input. Please enter a valid number between 1 and 301: ");
}
}
}
I ran this code, to see if it would show a better performance than yours.
Scanner sc = new Scanner(System.in);
boolean valid = true;
do {
if (!valid) {
System.out.print("Invalid Input. ");
}
System.out.print("Please enter a valid number between 1 and 301: ");
String input = sc.next();
try {
int value = Integer.parseInt(input);
valid = (value >= 1 && value <= 301);
} catch (NumberFormatException nfex) {
valid = false;
}
} while (!valid);
When the conversion to integer fails, the JVM hangs a little. I believe your problem has more to do with the try / catch mecanism that Scanner performs under the hood, than with design.
Assuming you want only 1 input from the user, try following simple code, which takes input from the user until user enters a valid input.
Scanner in = new Scanner(System.in);
int flag = 0,x=0;
while(flag == 0){
x = in.nextInt();
if(x<1 || x>301){
flag = 0;
System.out.println("Invalid Input.");
}
else{
flag = 1;
}
}
And if you want user to input more than 1 inputs (i.e 3 here), than set a counter that increases with every valid input of the user, as following:
Scanner in = new Scanner(System.in);
int flag = 0,x=0,count = 1;
while(flag == 0){
x = in.nextInt();
if(x<1 || x>301){
flag = 0;
System.out.println("Invalid Input.");
}
else{
//executes when input is valid
if(count == 3){
flag = 1;
}
count++;
}
}
Edit:
If you also want to check whether the input is Integer or not, than you have to add one extra condition in above code. And as you said you want only one input from user rather than 3, you have to change exit condition. Change code as following:
Scanner in = new Scanner(System.in);
int flag = 0,count = 1,x=0,flag1 = 0;
String y;
while(flag == 0){
y = in.next();
flag1 = 0;
try{
x = Integer.parseInt(y);
}
catch(NumberFormatException e){
flag1 = 1;
System.out.println("Invalid Input.");
}
if((x<1 || x>301)&&flag1 == 0){
flag = 0;
System.out.println("Invalid Input.");
}
else if(flag1 == 0){
//executes when input is valid
if(count == 1){ // put count == 3 if you want 3 inputs from user.
flag = 1;
}
count++;
}
}
Here we are taking the input as a String and than converting the String into the Integer by using Integer.parseInt(). If the String is not Integer, than it will throw the exception and we will continue the loop till the valid input is entered by the user.
Use DO WHILE for result
do{
System.out.print("value of x : " + x );
x++;
System.out.print("\n");
}while( x < 20 );
OK ?

How do I make sure User input integer type only and integer that is greater than 0?

I am attempting to make sure the user input int type only and make sure the integer inputted is greater than 0.
I was able to come up with the following to make sure the input is int type:
Scanner scan = new Scanner(System.in);
while(!scanner.hasNextInt())
{
scanner.next();
}
int input = scan.nextInt();
But how should I include a condition checking to make sure the integer is greater than 0 as well?
The problem with your current approach is you've already ready the value from the Scanner before it reaches int input = scan.nextInt();, meaning that by the time you use nextInt, there's nothing in the Scanner to be read and it will wait for the next input from user...
Instead, you could read the String from the Scanner using next, use Integer.parseInt to try and parse the result to an int and then check the result, for example...
Scanner scanner = new Scanner(System.in);
int intValue = -1;
do {
System.out.print("Please enter a integer value greater than 0: ");
String next = scanner.next();
try {
intValue = Integer.parseInt(next);
} catch (NumberFormatException exp) {
}
} while (intValue < 0);
System.out.println("You input " + intValue);
put an if statement inside your while loop like this
if(num <= 0){
System.out.println("Enter a number greater than zero");
}
else{
break;
}
You may use a condition in your code but not in the loop as.
`
Scanner scan = new Scanner(System.in);
abc:
while(!scanner.hasNextInt())
{
scanner.next();
}
int input = scan.nextInt();
if(input <= 0){
goto abc;
}
`
using .isDigit() method then checking to see if that number is greater than 0 if it is a digit

"Exception in thread" and more

import java.util.Scanner;
public class Power1Eng {
public static void main(String[] args) {
double x, prod = 1;
int n;
String s;
Scanner input = new Scanner(System.in);
System.out.print("This program prints x(x is a real number) raised to the power of n(n is an integer).\n");
outer_loop:
while (true) {
System.out.print("Input x and n: ");
x = input.nextDouble();
n = input.nextInt();
for (int i = 1; i <= n; i++) {
prod *= x;
}
System.out.printf("%.1f raised to the power of %d is %.4f. Do you want to continue?(Y/N) ", x, n, prod);
s = input.nextLine();
if (s.charAt(0) == 'Y')
continue;
else if (s.charAt(0) == 'N')
break;
else {
inner_loop:
while (true) {
System.out.print("Wrong input. Do you want to continue?(Y/N) ");
s = input.nextLine();
if (s.charAt(0) == 'Y')
continue outer_loop;
else if (s.charAt(0) == 'N')
break outer_loop;
else
continue inner_loop;
}
}
}
}
}
There was only trivial logical error when I used just next() method, but when I changed
next() method to nextLine() method, this error shows.
How can I fix this problem?
There are two problems. The first is that your string could be empty, and then fetching the first character will give an exception.
if (s.charAt(0) == 'Y') // This will throw if is empty.
Either test the length of the string to see if there is at least one character, or just use String.startsWith instead of charAt:
if (s.startsWith('Y'))
The second problem is that you entered a new line after your first input, and nextLine reads up to the next new line character only.
You could check for an initial character count, to make sure there are the correct number of characters that you expect. i.e:
while (true)
{
// ... some code ...
if (s.length() < 1)
{
continue;
}
// ... some code ...
}
This way, you wouldn't even have to continue running the rest of the code, which if the code base were larger, would help to optimize performance.
The "red text" that you see in the console is an indication of text being sent to standard error. In this case, it is an indication that your program has crashed.
The main problem you are encountering is with this logic:
System.out.print("Input x and n: ");
x = input.nextDouble();
n = input.nextInt();
for (int i = 1; i <= n; i++) {
prod *= x;
}
System.out.printf("%.1f raised to the power of %d is %.4f. Do you want to continue?(Y/N) ", x, n, prod);
s = input.nextLine();
Suppose the user input is:
2.1 4(enter)
input.nextDouble() will take 2.1, leaving 4(enter) on the standard input stream.
input.nextInt() will take 4, leaving (enter) on the standard input stream.
input.nextLine() will take "" (empty string), finally clearing the (enter) from the initial user input of x and n.

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