銀行卡號校驗位的LUHN演算法模10“隔位2倍加”校驗數的公式
計算LUHN演算法模10“隔位2倍加”校驗數的公式
計算步驟如下：
步驟1：從右邊第1個數字（低序）開始每隔一位乘以2。
步驟2：把在步驟1中獲得的乘積的各位數字與原號碼中未乘2的各位數字相加。
步驟3：從鄰近的較高的一個以0結尾的數中減去步驟2中所得到的總和［這相當於求這個總和的低位數字（個位數）的“10的補數”］。如果在步驟2得到的總和是以零結尾的數（如30、40等等），則校驗數字就是零。
例：
無校驗數字的卡號 4992 73 9871
4 9 9 2 7 3 9 8 7 1
×2 ×2 ×2 ×2 ×2
18 4 6 16 2
4+1+8+9+4+7+6+9+1+6+7+2=64
70-64=6
帶有校驗數字的卡號為：4992 73 9871 6
excel當中的校驗公式
=MOD(SUM(-MID(TEXT(MMULT((0&MID(LEFT(A2,LEN(A2)-1),ROW($1:$9)*2-MOD(LEN(A2)+{0,1},2),1))*{1,2},{100;1}),"000"),{1,2,3},1)),10)
陣列公式，要按shift+ctrl+enter結束
求魔方當中的公式應該怎樣寫呢？
(銀聯要求的卡號是13-19位此程式碼要滿足13-19位卡號的計算)
答覆：
完整支援16-19位銀號卡號校驗位計算：
text(mod((left(right($MF(賬號)$,3),1)+left(right($MF(賬號)$,5),1)+left(right($MF(賬號)$,7),1)+left(right($MF(賬號)$,9),1)+left(right($MF(賬號)$,11),1)+left(right($MF(賬號)$,13),1)+left(right($MF(賬號)$,15),1)+mod(left(right($MF(賬號)$,2),1)*2,10)+int(left(right($MF(賬號)$,2),1)*2/10)+mod(left(right($MF(賬號)$,4),1)*2,10)+int(left(right($MF(賬號)$,4),1)*2/10)+mod(left(right($MF(賬號)$,6),1)*2,10)+int(left(right($MF(賬號)$,6),1)*2/10)+mod(left(right($MF(賬號)$,8),1)*2,10)+int(left(right($MF(賬號)$,8),1)*2/10)+mod(left(right($MF(賬號)$,10),1)*2,10)+int(left(right($MF(賬號)$,10),1)*2/10)+mod(left(right($MF(賬號)$,12),1)*2,10)+int(left(right($MF(賬號)$,12),1)*2/10)+mod(left(right($MF(賬號)$,14),1)*2,10)+int(left(right($MF(賬號)$,14),1)*2/10)+mod(left(right($MF(賬號)$,16),1)*2,10)+int(left(right($MF(賬號)$,16),1)*2/10)+if(len($MF(賬號)$)>=17,left(right($MF(賬號)$,17),1),0)+if(len($MF(賬號)$)>=18,mod(value(left(right($MF(賬號)$,18),1))*2,10)+int(value(left(right($MF(賬號)$,18),1))*2/10),0)+if(len($MF(賬號)$)>=19,left(right($MF(賬號)$,19),1),0))*9,10),0)
配合銀行BIN碼的檢查，就可以知道銀行卡是出自哪個銀行。
功德圓滿
16-19位的校驗位公式
text(mod((left(right($MF(賬號)$,3),1)+left(right($MF(賬號)$,5),1)+left(right($MF(賬號)$,7),1)+left(right($MF(賬號)$,9),1)+left(right($MF(賬號)$,11),1)+left(right($MF(賬號)$,13),1)+left(right($MF(賬號)$,15),1)+mod(left(right($MF(賬號)$,2),1)*2,10)+int(left(right($MF(賬號)$,2),1)*2/10)+mod(left(right($MF(賬號)$,4),1)*2,10)+int(left(right($MF(賬號)$,4),1)*2/10)+mod(left(right($MF(賬號)$,6),1)*2,10)+int(left(right($MF(賬號)$,6),1)*2/10)+mod(left(right($MF(賬號)$,8),1)*2,10)+int(left(right($MF(賬號)$,8),1)*2/10)+mod(left(right($MF(賬號)$,10),1)*2,10)+int(left(right($MF(賬號)$,10),1)*2/10)+mod(left(right($MF(賬號)$,12),1)*2,10)+int(left(right($MF(賬號)$,12),1)*2/10)+mod(left(right($MF(賬號)$,14),1)*2,10)+int(left(right($MF(賬號)$,14),1)*2/10)+mod(left(right($MF(賬號)$,16),1)*2,10)+int(left(right($MF(賬號)$,16),1)*2/10)+if(len($MF(賬號)$)>=17,left(right($MF(賬號)$,17),1),0)+if(len($MF(賬號)$)>=18,mod(value(left(right($MF(賬號)$,18),1))*2,10)+int(value(left(right($MF(賬號)$,18),1))*2/10),0)+if(len($MF(賬號)$)>=19,left(right($MF(賬號)$,19),1),0))*9,10),0)
package com.cup.tracy;
public class Luhn {
/**
* Luhn演算法
* 根據卡號獲取校驗位
* @param cardNumber
* @return
*/
public static int getCheckNumber(String cardNumber){
int totalNumber = 0;
int lastNumber = 0;
for (int i = cardNumber.length()-1; i >= 0; i-=2) {
int tmpNumber = calculate(Integer.parseInt(String.valueOf(cardNumber.charAt(i))) * 2);
if (i==0) {
totalNumber += tmpNumber;
}else {
totalNumber += tmpNumber + Integer.parseInt(String.valueOf(cardNumber.charAt(i-1)));
}
}
if (totalNumber >= 0 && totalNumber < 9) {
return (10 - totalNumber);
}else {
String str = String.valueOf(totalNumber);
if (Integer.parseInt(String.valueOf(str.charAt(str.length()-1))) == 0) {
return 0;
}else {
return (10 - Integer.parseInt(String.valueOf(str.charAt(str.length()-1))));
}
}
}
/**
* 計算數字各位和
* @param number
* @return
*/
public static int calculate(int number){
String str = String.valueOf(number);
int total = 0;
for (int i = 0; i < str.length(); i++) {
total += Integer.valueOf(Integer.parseInt(String.valueOf(str.charAt(i))));
}
return total;
}
public static void main(String[] args) {
String cardNumber = "4992739871";
System.out.println(getCheckNumber(cardNumber));
}
}