How to give input array[] using scanner without space?

Question

In this below program, I'm trying to check whether the number is ISBN or not. I'm giving input with spaces (eg: 0 3 0 6 4 0 6 1 5 2) because array only accepts it like this. I don't know how to give input without space to read. Can anyone help me how to read the number eg: 0306406152 and also it will read 10 numbers only like if(i==10) else it says it's not ISBN number to give output.

public class ISBN {
 int digits[];
 int dig = 11;
 int sum;
 int isbn1;
 public void CheckISBN() {
 for (int digit : digits) {
 // System.out.println(digit);
 if (dig >= 1) {
 dig--;
 digit = digit * dig;
 // System.out.println(dig);
 }
 sum = sum + digit;
 isbn1 = sum % 11;
 }
 if (isbn1 == 0) {
 System.out.println(isbn1);
 System.out.println("it's valid ISBN number");
 } else {
 System.out.println("sorry it's not valid ISBN");
 }
 }
 public static void main(String[] args) {
 ISBN aa = new ISBN();
 aa.digits = new int[10];
 Scanner scan = new Scanner(System.in);
 int i = 0;
 while (scan.hasNextInt()) {
 aa.digits[i] = scan.nextInt();
 i++;
 if (i == 10) // aa.CheckISBN();
 {
 break;
 }
 for (int j = 0; j < aa.digits.length; j++) {
 // System.out.print(aa.digits[j]);
 }
 //System.out.println();
 }
 aa.CheckISBN();
 }
}

SAMPLE OUTPUT: 0 3 0 6 4 0 6 1 5 2
it's valid ISBN number

• Umair Farooq: why don't you take input in string rather than using array ?
• jrmullen: I don't understand the question. Your sample output and actual output are completely different. You don't even print the isbn in your output
• escape-llc: Scanner depends on whitespace to delimit tokens, so hasNextXXX works properly. Without spaces, your ISBN is basically a valid integer. Using a Pattern with a regex would be better for pattern-matching.

Answer 1

When the number is given without spaces,

import java.io.*;
import java.util.*;
public class ISBN {
 int digits[];
 int dig = 11;
 int sum;
 int isbn1;
 public void CheckISBN() {
 if(this.digits.length != 10)
 {
 System.out.println("sorry it's not valid ISBN");
 return;
 }
 for (int digit : digits) {
 // System.out.println(digit);
 if (dig >= 1) {
 dig--;
 digit = digit * dig;
 // System.out.println(dig);
 }
 sum = sum + digit;
 isbn1 = sum % 11;
 }
 if (isbn1 == 0) {
 //System.out.println(isbn1);
 System.out.println("it's valid ISBN number");
 } else {
 System.out.println("sorry it's not valid ISBN");
 }
 }
 public static void main(String[] args) {
 ISBN aa = new ISBN();
 Scanner scan = new Scanner(System.in);
 String num = scan.next(); //take input as a string
 int[] digits = new int[num.length()]; 
 for(int i = 0; i<digits.length; i++)
 digits[i] = num.charAt(i) - '0';
 aa.digits = digits;
 aa.CheckISBN();
 }
}

• pahan: You can accept the answer, if you think it answered your question
• KBM: yes, and can you explain this stmt please digits[i] = num.charAt(i) - '0';
• pahan: num is of type String. Strings are made of chars. We take the char at index i and convert it to an integer to put to digit array. How converstion occur is explained here.

Answer 2

Or scan it as int to get number format validation for free:

public class ISBN {
 public static void main(String[] args) {
 Scanner scan = new Scanner(System.in);
 if (scan.hasNextInt()) {
 checkISBN(scan.nextInt());
 }
 }
 public static void checkISBN(int isbn) {
 int sum = sum(digits(isbn));
 int isbn1 = sum % 11;
 if (isbn1 == 0) {
 System.out.println(isbn1);
 System.out.println("it's valid ISBN number");
 } else {
 System.out.println("sorry it's not valid ISBN");
 }
 }
 private static int sum(int[] digits) {
 return IntStream.rangeClosed(1, digits.length)
 .map(i -> i * digits[digits.length - i])
 .sum();
 }
 private static int[] digits(int isbn) {
 return Integer.toString(isbn)
 .chars()
 .map(c -> c - '0')
 .toArray();
 }
}

N.B.: It works for ISBN both with or without leading zeros.