信用卡验证 - 技术上正确但错误w /算法？

Question

因此，这是分配：使用8位数字 •从最右边的数字开始

信用卡号码检查，形成每一个其他数字的总和。例如，如果信用卡号码是4358 9795，那么您可以构成总和5 + 7 + 8 + 3 = 23. •将前面步骤中未包含的每个数字加倍。添加结果数字的所有数字。例如，使用上面给出的数字，将数字翻倍，从倒数第二位开始，产生18 18 10 8.将这些数值中的所有数字相加得到1 + 8 + 1 + 8 + 1 + 0 + 8 = 27. •添加前两个步骤的总和。如果结果的最后一位数字为0，则该数字有效。在我们的例子中，23 + 27 = 50，所以这个数字是有效的。 - 有效/无效？

我已经创建了一个测试程序和一个单独的方法。所有编译成功的应该不会有任何技术错误，但即使输入有效卡号时，程序也会返回该号码无效，因此我假设实际算法出现问题，但我很新这让我无法弄清楚它是什么。

import java.util.Scanner; 

public class CreditCardTester{ 

public static void main(String[] args){ 

    Scanner scanner = new Scanner(System.in); 
    String retry = ("y"); 
    String n = null; 

    while(retry.equalsIgnoreCase("y")){ // allows program to keep running even if the user enters in a capital Y 
     int lengthCheck = 1; 
     System.out.println("Please enter your 8-digit credit card number"); 

    // check to see whether the input number is exactly 8 digits 
     while(lengthCheck==1){ 
      n = scanner.next(); 
       if(n.length()==8) 
        lengthCheck=0; 
       else 
        System.out.println("Please make sure the credit card number you have entered is 8 digits"); 
     } // end inner while loop 

    // instantiate CardNumber and check validity 
    CardNumber number = new CardNumber(n); 
    number.check(); 
    if (number.isValid()) 
     System.out.println("The number you entered is a valid credit card number."); 
    else 
     System.out.println("The number you entered is not a valid credit card number. Would you like to enter in another number? (y/n)"); 
    retry = scanner.next(); 

    } // end outer while loop 

} 

} 

和单独的类

public class CardNumber { 

private String number; 
private boolean valid; 

public CardNumber(String n){ 
    number = n; 
    valid = true; 
} 

private void check1(){ 
    int a = Integer.parseInt(number.substring(7)); 
    int b = Integer.parseInt(number.substring(5,6)); 
    int c = Integer.parseInt(number.substring(3,4)); 
    int d = Integer.parseInt(number.substring(1,2)); 

    int oddsum = a + b + c + d; 

    int e = (Integer.parseInt(number.substring(6,7))) * 2; 
    int f = (Integer.parseInt(number.substring(4,5))) * 2; 
    int g = (Integer.parseInt(number.substring(2,3))) * 2; 
    int h = (Integer.parseInt(number.substring(0,1))) * 2; 

    String ee = String.valueOf(e); 
    String ff = String.valueOf(f); 
    String gg = String.valueOf(g); 
    String hh = String.valueOf(h); 

    int evensum = (Integer.parseInt(ee.substring(0))) + (Integer.parseInt(ff.substring(0))) + (Integer.parseInt(gg.substring(0))) + (Integer.parseInt(hh.substring(0))); 

    int totalsum = evensum + oddsum; 
    if (!(totalsum%10 == 0)) 
    valid=false; 
} 

public void check(){ 
    check1(); 
} 

public boolean isValid(){ 
    return valid; 
} 

} 

我敢肯定还有一个更好的方法来做到这一点，所以一切都是赞赏的建议！

Answer 1

信用卡号码验证称为Luhn算法。这里有一个java implementaion http://www.xinotes.org/notes/note/595/

为您的代码，我认为在这里：

int evensum = (Integer.parseInt(ee.substring(0))) + (Integer.parseInt(ff.substring(0))) + (Integer.parseInt(gg.substring(0))) + (Integer.parseInt(hh.substring(0))); 

你的意思是要总结两digist：

int evensum = Integer.parseInt(ee.substring(0,1)) + Integer.parseInt(ee.substring(1)) ... 

Answer 2

试试这个。

public class CardNumber { 

    String number; 
    boolean valid; 

    public CardNumber(String n){ 
     number = n; 
    } 

    public void check(){ 
     // The odd sum 
     // For the eight digits, 1, 3, 5, 7 are odd 
     int a = Integer.parseInt("" + number.charAt(1)); 
     int b = Integer.parseInt("" + number.charAt(3)); 
     int c = Integer.parseInt("" + number.charAt(5)); 
     int d = Integer.parseInt("" + number.charAt(7)); 

     int oddsum = a+b+c+d; 

     // The even sum 
     int e = (Integer.parseInt(number.substring(6,7))) * 2; 
     int f = (Integer.parseInt(number.substring(4,5))) * 2; 
     int g = (Integer.parseInt(number.substring(2,3))) * 2; 
     int h = (Integer.parseInt(number.substring(0,1))) * 2; 

     // As suggested by RUP to make it more simple 
     int evensum = e + f + g + h; 

     // Total sum 
     int totalsum = oddsum + evensum; 
     valid = (totalsum%10==0)?true:false;    
    } 

    public boolean isValid(){ 
     return valid; 
    } 

} 

Answer 3

你可以找到信用卡验证这里这是做两个信用卡号码验证加上信用卡类型检测的定制植入，

http://www.esupu.com/credit-card-validator-java/