用PHP验证信用卡的最佳方法是什么?


69

在没有信用卡号且没有其他信息的情况下,PHP中确定它是否为有效号码的最佳方法是什么?

现在,我需要可以与American Express,Discover,MasterCard和Visa一起使用的功能,但是如果它也可以与其他类型的功能一起使用,可能会有所帮助。

Answers:


150

卡号验证分为三个部分:

  1. 模式-是否与发行人模式(例如VISA /万事达卡等)匹配
  2. CHECKSUM-它是否实际进行校验和(例如,将“ 34”后的13个随机数字设为AMEX卡号)
  3. 真的存在-它实际上有一个相关的帐户(你不可能得到这个没有商家帐户)

模式

  • MASTERCARD前缀= 51-55,长度= 16(Mod10校验和)
  • VISA前缀= 4,长度= 13或16(Mod10)
  • AMEX前缀= 34或37,长度= 15(Mod10)
  • 大来俱乐部/卡特前缀= 300-305、36或38,长度= 14(Mod10)
  • 发现前缀= 6011,622126-622925,644-649,65,长度= 16,(Mod10)
  • 等(前缀的详细列表

校验和

大多数卡将Luhn算法用于校验和:

维基百科上描述的Luhn算法

Wikipedia链接上有许多实现的链接,包括PHP:

<?
/* Luhn algorithm number checker - (c) 2005-2008 shaman - www.planzero.org *
 * This code has been released into the public domain, however please      *
 * give credit to the original author where possible.                      */

function luhn_check($number) {

  // Strip any non-digits (useful for credit card numbers with spaces and hyphens)
  $number=preg_replace('/\D/', '', $number);

  // Set the string length and parity
  $number_length=strlen($number);
  $parity=$number_length % 2;

  // Loop through each digit and do the maths
  $total=0;
  for ($i=0; $i<$number_length; $i++) {
    $digit=$number[$i];
    // Multiply alternate digits by two
    if ($i % 2 == $parity) {
      $digit*=2;
      // If the sum is two digits, add them together (in effect)
      if ($digit > 9) {
        $digit-=9;
      }
    }
    // Total up the digits
    $total+=$digit;
  }

  // If the total mod 10 equals 0, the number is valid
  return ($total % 10 == 0) ? TRUE : FALSE;

}
?>

1
在模式下,您可以添加“发现”,前缀=“ 6”(也许是“ 60”),长度= 16
James Curran,

1
万事达具有51-55的前缀没有51或55根据beachnet.com/~hstiles/cardtype.html

8
此功能从字符串中删除所有非数字,因此卡号“ FRED”有效。调用此功能之前,请确保已验证卡号只有数字!
马特·康诺利

3
同样有效的信用卡号码也一样有效
-Gazillion

2
如文本中所述,@ BijuPDais-要检查它是否确实存在,您可能需要成为商家并实际尝试对卡进行计费。许多操作(例如酒店)都会收费,然后将少量金额退还到信用卡中。在所有验证方法中,这是唯一有效的验证卡是否有效的方法!
雷·海斯

30

10个正则表达式中,您不能没有PHP

function check_cc($cc, $extra_check = false){
    $cards = array(
        "visa" => "(4\d{12}(?:\d{3})?)",
        "amex" => "(3[47]\d{13})",
        "jcb" => "(35[2-8][89]\d\d\d{10})",
        "maestro" => "((?:5020|5038|6304|6579|6761)\d{12}(?:\d\d)?)",
        "solo" => "((?:6334|6767)\d{12}(?:\d\d)?\d?)",
        "mastercard" => "(5[1-5]\d{14})",
        "switch" => "(?:(?:(?:4903|4905|4911|4936|6333|6759)\d{12})|(?:(?:564182|633110)\d{10})(\d\d)?\d?)",
    );
    $names = array("Visa", "American Express", "JCB", "Maestro", "Solo", "Mastercard", "Switch");
    $matches = array();
    $pattern = "#^(?:".implode("|", $cards).")$#";
    $result = preg_match($pattern, str_replace(" ", "", $cc), $matches);
    if($extra_check && $result > 0){
        $result = (validatecard($cc))?1:0;
    }
    return ($result>0)?$names[sizeof($matches)-2]:false;
}

输入样例:

$cards = array(
    "4111 1111 1111 1111",
);

foreach($cards as $c){
    $check = check_cc($c, true);
    if($check!==false)
        echo $c." - ".$check;
    else
        echo "$c - Not a match";
    echo "<br/>";
}

这给了我们

4111 1111 1111 1111-签证

1
validatecard功能在哪里?
joseantgv


13

最好不要在代码末尾进行验证。将卡信息直接发送到您的付款网关,然后处理他们的回复。如果您不先执行Luhn检查之类的操作,它就可以帮助他们发现欺诈行为-让他们看到失败的尝试。


12
唯一的问题是每笔交易都有费用。它可能很小,但总和会增加,如果有人通过您的系统运行大量欺诈性卡号,则费用可能会不菲。
约翰·孔德

6

PHP代码

function validateCC($cc_num, $type) {

    if($type == "American") {
    $denum = "American Express";
    } elseif($type == "Dinners") {
    $denum = "Diner's Club";
    } elseif($type == "Discover") {
    $denum = "Discover";
    } elseif($type == "Master") {
    $denum = "Master Card";
    } elseif($type == "Visa") {
    $denum = "Visa";
    }

    if($type == "American") {
    $pattern = "/^([34|37]{2})([0-9]{13})$/";//American Express
    if (preg_match($pattern,$cc_num)) {
    $verified = true;
    } else {
    $verified = false;
    }


    } elseif($type == "Dinners") {
    $pattern = "/^([30|36|38]{2})([0-9]{12})$/";//Diner's Club
    if (preg_match($pattern,$cc_num)) {
    $verified = true;
    } else {
    $verified = false;
    }


    } elseif($type == "Discover") {
    $pattern = "/^([6011]{4})([0-9]{12})$/";//Discover Card
    if (preg_match($pattern,$cc_num)) {
    $verified = true;
    } else {
    $verified = false;
    }


    } elseif($type == "Master") {
    $pattern = "/^([51|52|53|54|55]{2})([0-9]{14})$/";//Mastercard
    if (preg_match($pattern,$cc_num)) {
    $verified = true;
    } else {
    $verified = false;
    }


    } elseif($type == "Visa") {
    $pattern = "/^([4]{1})([0-9]{12,15})$/";//Visa
    if (preg_match($pattern,$cc_num)) {
    $verified = true;
    } else {
    $verified = false;
    }

    }

    if($verified == false) {
    //Do something here in case the validation fails
    echo "Credit card invalid. Please make sure that you entered a valid <em>" . $denum . "</em> credit card ";

    } else { //if it will pass...do something
    echo "Your <em>" . $denum . "</em> credit card is valid";
    }


}

用法

echo validateCC("1738292928284637", "Dinners");

更多理论信息可以在这里找到:

信用卡验证-校验数字

校验和


1
该算法可能会大致识别出卡发行者,它不会验证数字是否合理..例如,它不会使用Luhn算法检查校验和!
Ray Hayes



0

只是抛出一些其他人可能会觉得有用的代码片段(不是PHP代码)。

PYTHON(单行代码;可能效率不高)

验证:

>>> not(sum(map(int, ''.join(str(n*(i%2+1)) for i, n in enumerate(map(int, reversed('1234567890123452'))))))%10)
True
>>> not(sum(map(int, ''.join(str(n*(i%2+1)) for i, n in enumerate(map(int, reversed('1234567890123451'))))))%10)
False

要返回所需的校验位:

>>> (10-sum(map(int, ''.join(str(n*(i%2+1)) for i, n in enumerate(map(int, reversed('123456789012345')), start=1)))))%10
2
>>> (10-sum(map(int, ''.join(str(n*(i%2+1)) for i, n in enumerate(map(int, reversed('234567890123451')), start=1)))))%10
1

MySQL函数

函数“ ccc”和“ ccd”(信用卡检查和信用卡数字)

请注意,“ ccc”函数还要进行额外的检查,如果计算得出的总和为0,则返回的结果将始终为FALSE,因此,全零的CC编号将永远不会被验证为正确的(在正常情况下,它将被正确验证)。可以根据需要添加/删除此功能。可能有用,具体取决于特定要求。

DROP FUNCTION IF EXISTS ccc;
DROP FUNCTION IF EXISTS ccd;

DELIMITER //

CREATE FUNCTION ccc (n TINYTEXT) RETURNS BOOL
BEGIN
  DECLARE x TINYINT UNSIGNED;
  DECLARE l TINYINT UNSIGNED DEFAULT length(n);
  DECLARE i TINYINT UNSIGNED DEFAULT l;
  DECLARE s SMALLINT UNSIGNED DEFAULT 0;
  WHILE i > 0 DO
    SET x = mid(n,i,1);
    IF (l-i) mod 2 = 1 THEN
      SET x = x * 2;
    END IF;
    SET s = s + x div 10 + x mod 10;
    SET i = i - 1;
  END WHILE;
  RETURN s != 0 && s mod 10 = 0;
END;

CREATE FUNCTION ccd (n TINYTEXT) RETURNS TINYINT
BEGIN
  DECLARE x TINYINT UNSIGNED;
  DECLARE l TINYINT UNSIGNED DEFAULT length(n);
  DECLARE i TINYINT UNSIGNED DEFAULT l;
  DECLARE s SMALLINT UNSIGNED DEFAULT 0;
  WHILE i > 0 DO
    SET x = mid(n,i,1);
    IF (l-i) mod 2 = 0 THEN
      SET x = x * 2;
    END IF;
    SET s = s + x div 10 + x mod 10;
    SET i = i - 1;
  END WHILE;
  RETURN ceil(s/10)*10-s;
END;

然后可以在SQL查询中直接使用函数:

mysql> SELECT ccc(1234567890123452);
+-----------------------+
| ccc(1234567890123452) |
+-----------------------+
|                     1 |
+-----------------------+
1 row in set (0.00 sec)

mysql> SELECT ccc(1234567890123451);
+-----------------------+
| ccc(1234567890123451) |
+-----------------------+
|                     0 |
+-----------------------+
1 row in set (0.00 sec)

mysql> SELECT ccd(123456789012345);
+----------------------+
| ccd(123456789012345) |
+----------------------+
|                    2 |
+----------------------+
1 row in set (0.00 sec)

mysql> SELECT ccd(234567890123451);
+----------------------+
| ccd(234567890123451) |
+----------------------+
|                    1 |
+----------------------+
1 row in set (0.00 sec)

By using our site, you acknowledge that you have read and understand our Cookie Policy and Privacy Policy.
Licensed under cc by-sa 3.0 with attribution required.