1

所以这是任务：

信用卡号码 使用 8 位数字检查 • 从最右边的数字开始，形成所有其他数字的总和。例如，如果信用卡号是 4358 9795，那么您将形成总和 5 + 7 + 8 + 3 = 23。 • 将前面步骤中未包括的每个数字加倍。添加结果数字的所有数字。例如，对于上面给出的数字，将数字加倍，从倒数第二个开始，得到 18 18 10 8。将这些值中的所有数字相加得到 1 + 8 + 1 + 8 + 1 + 0 + 8 = 27。 • 将前面两个步骤的总和相加。如果结果的最后一位为 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？

我已经为这些方法创建了一个测试程序和一个单独的程序。一切都成功编译，所以不应该有任何技术错误，但即使我输入有效的卡号，程序也会返回该号码无效，所以我假设实际算法有问题，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

}

}

和单独的班级

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;
}

}

我相信还有一个更好的方法可以做到这一点，所以所有的建议都值得赞赏！

4

3 回答 3

1

The credit card number validation is called the Luhn Algorithm. Here's a java implementaion http://www.xinotes.org/notes/note/595/

For your code I think here:

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

You mean to sum both digist: 

int evensum = Integer.parseInt(ee.substring(0,1)) +  Integer.parseInt(ee.substring(1)) ...
于 2012-10-05T01:12:02.063 回答
0

您可以在此处找到信用卡验证器的自定义植入，它同时进行信用卡号验证和信用卡类型检测，

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

于 2013-10-23T18:17:00.347 回答
0

尝试这个。

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;
    }

}
于 2012-10-05T01:28:37.203 回答