java - Luhn 检查 Java 中的程序计算不正确

Question

我用 Java 创建了这个 Luhn 检查(或 Mod 10 检查)，偶数和奇数总和没有正确相加，我不明白为什么。当我单独写出这一节时，它起作用了，而且看起来效果很好。作为一个与所有其他方法一起使用的整个程序，它不起作用。大家有什么想法吗？

输入号码 4388576018410707 应该是有效的。

import java.util.Scanner;
import java.io.*;

public class combineAll {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.print("Enter a credit card number: ");
        long userInput = input.nextLong();

        int getSize=getSize(userInput); //Run getSize() Method.
        Integer z = (int) (long) getPrefix(userInput, getSize); //Run getPrefix() Method.

        if (prefixMatch(userInput, z)== true) { //Run prefixMatch() Method.
            long n=sumbOfDoubleEvenPlace(userInput); //Run sumbOfDoubleEvenPlace() Method.
            long m=sumOfOddPlace(userInput);  //Run sumOfOddPlace() Method.
            System.out.println("Total Even: " +n);
            System.out.println("Total Odd: " +m);
            long v=n+m;
            if (isValid(v)==true) {
                System.out.println("Valid");
            } else if (isValid(v)==false){
                System.out.println("Invalid");
            }
        } else {
            System.out.println("Invalid");
        }
    } //End main

    //Return the number of digits in d
    public static int getSize(long d) {
        String str = Long.toString(d);
        int x = str.length();
        return x;
    }

    //Return the first k number of digits from number. If the number of digits in number is less than k, return number
    public static long getPrefix(long number, int k) {
        int z=0;
        if (k>=13 && k<=16) {
            String str = Long.toString(number);
            String g = str.substring(0,1);
            String h = str.substring(0,2);

            int d=Integer.parseInt(g);
            int q=Integer.parseInt(h);

            if (d==4 || d==5 ||d==6) {
                z=d;
            } else if (q==37) {
                z=q;
            }

        } else {
            z=-1;
        }

        return z;
    }

    //Return true if the digit d is a prefix for number
    public static boolean prefixMatch(long number, int d) {
        if (d==4 || d==5 || d==6 || d==37) {
            return true;
        } else {
            return false;
        } 
    }

    //Get the result from step 2
    public static int sumbOfDoubleEvenPlace(long number) {
        long a=number;
        int d=0; //Adds each individual numbers.
        while (a>0) {
            long b=0;
            long c=0; //Equals the Mod of UserInput.
            a=a/10;
            c=a%10;
            System.out.println("even: " +c);
            b=c*2;

            if (b>=10) {
                Integer digit = (int) (long) b;
                d+=getDigit(digit); //Run getDigit() Method.
            } else {
                Integer digit = (int) (long) b;
                d+=b;
            }

            a=a/10; //Advance decimal one space to the left.
        }
        return d;
    }

    //Return sum of odd-place digits in number
    public static int sumOfOddPlace(long number) {
        long a=number;
        int d=0; //Adds each individual numbers.
        while (a>0) {
            long b=0;
            long c=0; //Equals the Mod of UserInput.
            c=a%10;
            System.out.println("odd: " +c); //Print for debugging.
            b=c*2;

            if (b>=10) {
                Integer digit = (int) (long) b;
                d+=getDigit(digit); //Run getDigit() Method.
            } else {
                Integer digit = (int) (long) b;
                d+=b;
            }

            a=a/10; //Advance decimal one space to the left.
            a=a/10;
        }
        return d;
    }

    //Return this number if it is a single digit, otherwise return the sum of the two digits
    public static int getDigit(int number) {
        int d=0;
        int x=0;
        int y=number;
        while (y>0) {
            x+=y%10;
            y=y/10;
        }
        return x;
    }

    //Return true if the card number is valid
    public static boolean isValid(long number) {    
        long c=number%10;
        if (c==0) {
            return true;
        } else {
            return false;
        }
    }
}

Answer 1

代码有点难以理解。基于算法描述形式Wikipedia实现应该简单得多。 Here是遵循 Wikipedia 的描述的其他实现.

public class Cards {
/**
 * Checks if the card is valid
 * 
 * @param card
 *           {@link String} card number
 * @return result {@link boolean} true of false
 */
public static boolean luhnCheck(String card) {
    if (card == null)
        return false;
    char checkDigit = card.charAt(card.length() - 1);
    String digit = calculateCheckDigit(card.substring(0, card.length() - 1));
    return checkDigit == digit.charAt(0);
}

/**
 * Calculates the last digits for the card number received as parameter
 * 
 * @param card
 *           {@link String} number
 * @return {@link String} the check digit
 */
public static String calculateCheckDigit(String card) {
    if (card == null)
        return null;
    String digit;
    /* convert to array of int for simplicity */
    int[] digits = new int[card.length()];
    for (int i = 0; i < card.length(); i++) {
        digits[i] = Character.getNumericValue(card.charAt(i));
    }

    /* double every other starting from right - jumping from 2 in 2 */
    for (int i = digits.length - 1; i >= 0; i -= 2)    {
        digits[i] += digits[i];

        /* taking the sum of digits grater than 10 - simple trick by substract 9 */
        if (digits[i] >= 10) {
            digits[i] = digits[i] - 9;
        }
    }
    int sum = 0;
    for (int i = 0; i < digits.length; i++) {
        sum += digits[i];
    }
    /* multiply by 9 step */
    sum = sum * 9;

    /* convert to string to be easier to take the last digit */
    digit = sum + "";
    return digit.substring(digit.length() - 1);
}

public static void main(String[] args)    {
    String pan = "4388576018410707";
    System.out.println("Validate pan number '" + pan + "': " + luhnCheck(pan2));
}    
}