Question

我是完成程序的最后一步，但每当我输入整数（长整数）时，我输入输入不匹配：

Compiler message: "Exception in thread "main" java.util.InputMismatchException: For input               string: "4388576018402626"
at java.util.Scanner.nextInt(Scanner.java:2097)
at java.util.Scanner.nextInt(Scanner.java:2050)
at CreditCardValidation.main(CreditCardValidation.java:12)"

我的代码如下：

import java.util.Scanner ; //import Scanner

public class CreditCardValidation {

public static void main (String[] args){

    Scanner kbd = new Scanner(System.in) ;

    System.out.println("Please enter Creditcard number: " ) ;

    int credNumber = kbd.nextInt() ;

    boolean n = isValid( credNumber ) ;

if (credNumber == 0 )
    System.exit(0) ;

do {                

    System.out.println("Please enter Creditcard number: " ) ;
    credNumber = kbd.nextInt() ;
    }
    while ( credNumber < 0 ) ;

if (credNumber > 0 )
    System.out.print("This credit card number is " + n ) ;




}

/*
Return true is the number is a valid card number.
*/

public static boolean isValid(long number) {


    long p = getPrefix(number, 1);
    long p2 = getPrefix(number, 2);
    int n = getSize(number);


    if ((p == 4 || p == 5 || p == 6 || p2 == 37)&& (n < 13 || n > 16) && (((sumOfDoubleEvenPlace(number) + sumOfoddPlace(number))) % 10) == 0)
        return true ;
    else
        return false ;

}


/* The sum of every other digit, doubled, starting with the first digit. */

public static int sumOfDoubleEvenPlace(long number) {
    int sum = 0;
    int maxDigitLenth = 16;
    for (int i = 1; i <= maxDigitLenth; i++)
    {
        if (i % 2 == 0)
        {
            sum = sum + getDigit((int)(number % 10) * 2);
        }
        number /= 10;
    }
    return sum;
}

/*
Return the number if it is 0-9, otherwise return the sum of
the digits of the number.
*/

public static int getDigit(int number) {
    if (number < 10) {

        return number;
    }
    else {

        return (number / 10) + (number % 10);
    }

    }
/*
Return the sum of the odd-place digits.
*/

public static int sumOfoddPlace(long number) {
    int maxDigitLength = 16;
    int sum = 0;
    for (int i = 1; i <= maxDigitLength; i++)
    {

        if (i % 2 == 1)
        {
            sum = sum + (int)(number % 10);
        }
       number /= 10;
    }
    return sum;
}

/*
Return the number of digits in d
*/

public static int getSize(long d) {

    int size = 0 ;

    while( d > 0 ) {
        d = d / 10 ;
        size = size + 1 ;       
    }
    return size ;
}

/*
Return the first k number of digits from number. If the number of digits in number is 
less than k, return the number.
*/

public static long getPrefix(long n, int k) {

    int f = getSize(n)-k;

    long prefix = n/((long)(Math.pow(10, f)));

    return prefix;
    }

/*
Return true if the digit d is a prefix for number.
*/

public static boolean prefixMatched( long number, int d ) {

    if ( d == getPrefix(number, 4))

        return true ;
    else
        return false ;

    }

}

感谢您的时间！

Answer 1

最大整数值为2147483647.而是使用long：

long credNumber = kbd.nextLong();

或更好地使用字符串作为信用卡号：

String credNumber = kbd.nextLine();

Answer 2

这是因为您输入的值超出了整数值的范围。在这种情况下你需要长时间使用。整数的最大值是 2147483647 。

long credNumber = kbd.nextLong();
..
// in the do while loop also
credNumber = kbd.nextLong() ;

Answer 3

您应该使用：

long credNumber = kbd.nextLong();
boolean n = isValid(credNumber);

因为您输入的值（16位数）超出了int的限制。

您的验证方法接受的时间很长，因此您的代码应该可以正常运行以上更改。

public static boolean isValid(long number)

Answer 4

您也可以将int转换为BigInteger，然后在BigInteger对象上调用add函数

扫描程序用于长整数，线程“main”中的异常java.util.InputMismatchException

4 个答案: