java - 无法找出错误 Luhn 检查

Question

它应该使用 luhn 检查来告诉我一张卡是有效还是无效4388576018402626 无效4388576018410707 有效但它一直告诉我一切都是无效的:/任何关于做什么或去哪里寻找的建议都会很棒。我已经被困了几个小时了。如果人们告诉我有关如何查找代码未按预期工作的原因的任何提示，也会有所帮助。我使用 eclipse 和 java

public class Task11 {

   public static void main(String[] args) {
        Scanner input = new Scanner(System.in);

        System.out.print("Enter a credit card number as a long integer: ");
        long number = input.nextLong();


        if (isValid(number)) {
            System.out.println(number + " is valid");
        } else {
            System.out.println(number + " is invalid");
        }


    }

    public static boolean isValid(long number) {

        return (getSize(number) >= 13) && (getSize(number) <= 16)
                && (prefixMatched(number, 4) || prefixMatched(number, 5) || prefixMatched(number, 6) || prefixMatched(number, 37))
                && (sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 == 0;
    }

    public static int sumOfDoubleEvenPlace(long number) {

        int result = 0;

        long start = 0;

        String digits = Long.toString(number);

        if ((digits.length() % 2) == 0) {
            start = digits.length() - 1;
        } else {
            start = digits.length() - 2;
        }


        while (start != 0) {


            result += (int) ((((start % 10) * 2) % 10) + (((start % 10) * 2) / 2));

            start = start / 100;

        }

        return result;
    }


    public static int getDigit(int number) {

        return number % 10 + (number / 10);

    }

    public static int sumOfOddPlace(long number) {

        int result = 0;

        while (number != 0) {

            result += (int) (number % 10);

            number = number / 100;

        }

        return result;

    }

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

        return getPrefix(number, getSize(d)) == d;

    }

    public static int getSize(long d) {

        int numberOfDigits = 0;

        String sizeString = Long.toString(d);
        numberOfDigits = sizeString.length();

        return numberOfDigits;

    }


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

        String size = Long.toString(number);

        if (size.length() <= k) {
            return number;
        } else {
            return Long.parseLong(size.substring(0, k));
        }
    }
}

Answer 1

您应该修改您的isValid()当它不起作用时写下的方法，如下所示:

 public static boolean isValid(long number) {
        System.err.println();
        if(getSize(number) < 13){
            System.out.println("Err: Number "+number+" is too short");
            return false;
        } else if (getSize(number) > 16){
  public static boolean isValid(long number) {
        System.err.println();
        if(getSize(number) < 13){
            System.out.println("Err: Number "+number+" is too short");
            return false;
        } else if (getSize(number) > 16){
            System.out.println("Err: Number "+number+" is too long");
            return false;
        } else if (! (prefixMatched(number, 4) || prefixMatched(number, 5) || prefixMatched(number, 6) || prefixMatched(number, 37)) ){
            System.out.println("Err: Number "+number+" prefix doesn't match");
            return false;
        } else if( (sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 != 0){
            System.out.println("Err: Number "+number+" doesn't have sum of odd and evens % 10. ");
            return false;
        }  
        return true;
    }

我对你的问题的猜测是在getPrefix()上方法，你也应该在这里添加一些日志。

编辑:所以，有更多时间来帮助您(不知道是否仍然有必要，但无论如何)。另外，我纠正了我写的方法，有一些错误(例如， getSize(number) >= 13 的相反是 getSize(number) < 13 )...首先，使用一组数据进行测试会比每次自己输入值更快(添加您要检查的值):

public static void main(String[] args) {
           long[] luhnCheckSet = {
                   0, // too short
                   1111111111111111111L, // too long (19)
                   222222222222222l // prefix doesn't match
                   4388576018402626l, // should work ?
                   };

        //System.out.print("Enter a credit card number as a long integer: ");
        //long number = input.nextLong();

        for(long number : luhnCheckSet){
            System.out.println("Checking number: "+number);
            if (isValid(number)) {
                System.out.println(number + " is valid");
            } else {
                System.out.println(number + " is invalid");
            }
            System.out.println("-");
        }
    }

我不知道这件事的细节，但我认为你应该与 String 合作一直以来，仅在需要时解析为 long (如果数字超过 19 个字符，它可能不会解析它 long )。尽管如此，还是做多。

我详细介绍了您的getPrefix()更多日志并放入 d in 参数 in long (小心比较原始类型是个好习惯):

public static boolean prefixMatched(long number, long d) {
            int prefixSize = getSize(d);
            long numberPrefix = getPrefix(number, prefixSize);
            System.out.println("Testing prefix of size "+prefixSize+" from number: "+number+". Prefix is: "+numberPrefix+", should be:"+d+", are they equals ? "+(numberPrefix == d));
            return numberPrefix == d;
        }

仍然不知道这段代码有什么问题，但看起来它来自上次测试:我没有这样做，但你应该从 sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 中创建一种方法并记录数字和总和(就像我在 prefixMatched() 中所做的那样)。在这两种方法中添加日志，以确保它获得您想要的结果/正常工作。你用过调试器吗？如果可以的话，就这样做，它比添加大量日志要快!

祝你好运