Java中的Luhn公式实现

Posted 2023-02-22

【中文标题】Java中的Luhn公式实现

【英文标题】：Luhn formula implementation in Java

【发布时间】：2015-05-26 02:49:57

【问题描述】：

我正在尝试在我的 java servlet 应用程序中实现一个 luhn 公式。我尝试了其他散布在互联网上的“有效”信用卡号码，但没有用。我只是想知道我是否正确理解了它。任何帮助将不胜感激！

    public static boolean luhn(String input)
         char[] creditCard  = input.toCharArray();
         int checkSum = 0;
         boolean alternate = false;

         for (int i = creditCard.length - 1; i >= 0; i --)
              int m = (int)Integer.parseInt(Character.toString(creditCard[i]));
              if (alternate)
                  m *= 2;
                  if (m > 9)
                     m = (m & 10) + 1;
                  
             
        checkSum += m;
        alternate = true;
    

    if ( (checkSum % 10) == 0)
        return true;
    else
        return false;
    

【参考方案1】：

这里是工作代码

public static void main(String[] args) 
Scanner in = new Scanner(System.in);
boolean repeat;
List<Integer> digits = new ArrayList<Integer>();

do 
    repeat = false;
    System.out.print("Enter your Credit Card Number : ");
    String input = in.next();

    for (int i = 0; i < input.length(); i++) 
        char c = input.charAt(i);
        if (c < '0' || c > '9') 
            repeat = true;
            digits.clear();
            break;
         else 
            digits.add(Integer.valueOf(c - '0'));
        
    
 while (repeat);

int[] array = new int[digits.size()];
for (int i = 0; i < array.length; i++) 
    array[i] = Integer.valueOf(digits.get(i));

boolean valid = check(array);
System.out.println("Valid: " + valid);

检查 luhn 算法

public static boolean check(int[] digits) 
 int sum = 0;
 int length = digits.length;
 for (int i = 0; i < length; i++) 

   // get digits in reverse order
   int digit = digits[length - i - 1];

   // every 2nd number multiply with 2
   if (i % 2 == 1) 
       digit *= 2;
   
   sum += digit > 9 ? digit - 9 : digit;
 
 return sum % 10 == 0;

或者更复杂的程序可能如下

import java.util.Scanner;
public class Luhn 
private static Scanner input;

public static void main(String... args) 
input = new Scanner(System.in);
System.out.print("Enter number to validate:\n");
String pnr = input.nextLine();
boolean result = luhn(pnr);
printMessage(result);
input.close();


static boolean luhn(String pnr)
// this only works if you are certain all input will be at least 10   characters
int extraChars = pnr.length() - 10;
if (extraChars < 0) 
  throw new IllegalArgumentException("Number length must be at least 10 characters!");

pnr = pnr.substring(extraChars, 10 + extraChars);
int sum = 0;
for (int i = 0; i < pnr.length(); i++)
  char tmp = pnr.charAt(i);
  int num = tmp - '0';
   int product;
   if (i % 2 != 0)
     product = num * 1;
   
   else
    product = num * 2;
  
   if (product > 9)
    product -= 9;
   sum+= product;              
 
  return (sum % 10 == 0);
 

 private static void printMessage(boolean valid) 
  if (valid)
  System.out.print("Valid!\r");
  
  else
  System.out.print("Invalid!");
 
 

【讨论】：

感谢您的帮助。我很感激！我只是想知道我的代码是否是信用卡有效性的真正检查器。

@user3323654 您的代码错过了alternate 的更改状态，因此它不会起作用，更好地理解会更麻烦，您可以使用上面提到的代码:)，如果它帮助您将其标记为接受:)