Luhn's algorithm in php

gistfile1.aw

<?php
function is_valid_luhn($number) {
  settype($number, 'string');
  $sumTable = array(
    array(0,1,2,3,4,5,6,7,8,9),
    array(0,2,4,6,8,1,3,5,7,9));
  $sum = 0;
  $flip = 0;
  for ($i = strlen($number) - 1; $i >= 0; $i--) {
    $sum += $sumTable[$flip++ & 0x1][$number[$i]];
  }
  return $sum % 10 === 0;
}

9) { $val -=9; } $sum += $val; } return(($sum % 10) == 0); } if (! is_valid_nums($_POST['nums'])) { echo 'the num is invalid.'; } } ?>

Please be aware that this functions will only work if the passed argument is a string, else it will fail.
For more info read:

http://lamp-ios.com/?p=256

It has to do with 1 thing:

1. Your platform (32 or 64 bit)

Be aware of this

Here is a simple implementation:

/**
 * @see http://en.wikipedia.org/wiki/Luhn_algorithm
 */
private function checksum ($card_number) {
    $card_number_checksum = '';

    foreach (str_split(strrev((string) $card_number)) as $i => $d) {
        $card_number_checksum .= $i %2 !== 0 ? $d * 2 : $d;
    }

    return array_sum(str_split($card_number_checksum)) % 10 === 0;
}

You can simply typecast whatever input you have into a string before applying it to the function or you could incorporate a step to first typecast your arguments into strings.

This function works perfectly if your number is already in a 16-character string, is_valid_luhn("4111111111111111")

But the function will fail if the string contains any symbols or spaces, is_valid_luhn("4111 1111 1111 1111") = false

This can be easily fixed by placing a preg_replace at the beginning of the function to strip all non-digit characters.

$number = preg_replace("/[^0-9]/", "", $number);

PHP:

function isValid($num) {
    $num = preg_replace('/[^\d]/', '', $num);
    $sum = '';

    for ($i = strlen($num) - 1; $i >= 0; -- $i) {
        $sum .= $i & 1 ? $num[$i] : $num[$i] * 2;
    }

    return array_sum(str_split($sum)) % 10 === 0;
}

Javascript:

function isValid(number) {
    var num = number.replace(/[^\d]/, '');
    var str = '';

    for (var i = num.length - 1; i >= 0; -- i) {
        str += i & 1 ? num[i] : (parseInt(num[i]) * 2).toString();
    }

    var sum = str.split('').reduce(function(prev, current) {
        return prev + parseInt(current);
    }, 0);

    return sum % 10 === 0;
};

= 0; $i--) { $sum += $sumTable[$flip++ & 0x1][$number[$i]]; } return $sum % 10 === 0; }

        //Using Luhn's algorithm to validate a credit card

        public static function isValidCardNumber($number){
        	settype($number, 'string');
        	$number = preg_replace("/[^0-9]/", "", $number);
        	$numberChecksum= '';
		$reversedNumberArray = str_split(strrev($number));
		foreach ($reversedNumberArray as $i => $d) {
		    $numberChecksum.= (($i % 2) !== 0) ? (string)((int)$d * 2) : $d;
		}
		$sum = array_sum(str_split($numberChecksum));
                return ($sum % 10) === 0;
        }

is there any possibility to get bpay biller address and info by biller code?

@mikemirten first I thought thats really small and simple to understand but unfortunately it's wrong. have you tried this with a credit card number of odd length? visas can have 13 - 16 numbers and that fails.

my solution is now:

function validateLuhn(string $number): bool
{
    $sum = 0;
    $revNumber = strrev($number);
    $len = strlen($number);

    for ($i = 0; $i < $len; $i++) {
        $sum += $i & 1 ? ($revNumber[$i] > 4 ? $revNumber[$i] * 2 - 9 : $revNumber[$i] * 2) : $revNumber[$i];
    }

    return $sum % 10 === 0;
}

@mogzol: you are right: the string sum does not make sense... So I removed it.

edit fixed the calculation without string split and unused variables

Alternative version of @tflori's code that doesn't rely on strrev, array_sum or str_split, which makes it more readable IMO, and slightly faster (though at this point it's basically just the original code without the sum table):

function validateLuhn(string $number): bool
{
    $sum = 0;
    $flag = 0;

    for ($i = strlen($number) - 1; $i >= 0; $i--) {
        $add = $flag++ & 1 ? $number[$i] * 2 : $number[$i];
        $sum += $add > 9 ? $add - 9 : $add;
    }

    return $sum % 10 === 0;
}

@tflori Your updated code doesn't work just FYI. You actually did need the array_sum. This is because of the first step outlined here: https://en.wikipedia.org/wiki/Luhn_algorithm#Description

If the result of this doubling operation is greater than 9 (e.g., 8 × 2 = 16), then add the digits of the result (e.g., 16: 1 + 6 = 7, 18: 1 + 8 = 9)

Your original code achieved this by using a string and summing all the numbers in the string. My code uses the alternate method mentioned:

alternatively, the same final result can be found by subtracting 9 from that result (e.g., 16: 16 − 9 = 7, 18: 18 − 9 = 9)

But your newly edited code does neither, and as such returns false for many valid numbers.

@mogzol ah f**k.. thanks for the info 👍

In case anyone tries using @mogzol solution while working with the Luhn algorithm applied to a base higher than 10, for example to validate an alphanumeric string where each character can go from 0-9 or A-Z, meaning its value ranges from 0-36, you need to be more careful performing the total sum.

In the solution, when we're doing the total sum we are essentially ignoring the fact that Luhn algorithm only wants you to subtract 9 to values that have been doubled. It works for base 10 only because values that have not been doubled can't possibly be greater than 9.

To correctly implement a general any-base Luhn algorithm, you don't have this luxury, and must therefore apply both the doubling of the value and subsequent subtraction only IF the condition is met, in this case ($flag++ & 1), which by the way is quite an unreadable way to flip between true and false.

Here's a general readable example, assuming some valid string $number in base $base

$flip = false;
    $sum = 0;
    for ($i = strlen($number) - 1; $i >= 0; $i--) {
        $value = intval($number[$i], $base);
        if($flip){
            $value = $value * 2;
            $value = $value > 9 ? $value - 9 : $value;
        }
        $sum += $value;
        $flip = !$flip;
    }
    return $sum % 10 === 0;