How to Calculate and Get Digit Using Module 11 Algorithm - php

Question

I'm trying to do a syntax cast where I have a javascript function that calculates the check digit from an input number.
where the variables:
input - is the input value (Example: 200300)
num_digits - is the number of digits (if defined 1, its respective digit will be 7; already defined it 2, its respective digit will be 70. According to the input value)
limit - is the multiplication limit (in my case I need it to be multiplied by/up to 9)
x10 - in this case being true or false, being true the digit will be multiplied by 10
all variables mentioned above refer to my JavaScript function:
function calcDigitMod11(input, num_digits, limit, x10) {
var mult, sum, i, n, digit;
if (!x10) num_digits = 1;
for (n = 1; n <= num_digits; n++) {
sum = 0; mult = 2;
for (i = (input.length - 1); i >= 0; i--) {
sum += (mult * parseInt(input.charAt(i)));
if (++mult > limit) mult = 2;
}
if (x10) {
digit = ((sum * 10) % 11) % 10;
} else {
digit = sum % 11;
if (digit == 10) digit = 'x';
}
input += (digit);
}
return input.substring((input.length - num_digits), num_digits);
}
in short, my big problem is to create this same function in the php syntax
If you print the function's return on the console, passing the following parameters: calcDigitMod11(200300, 1, 9, true); your return must be the check digit
7
giving an applied...
"tried" to convert syntaxes using the same variables, parameters, among others
function calcDigitMod11($input, $num_digits, $limit, $x10) {
$mult; $sum; $i; $n; $digit;
if (!$x10) $num_digits = 1;
for ($n = 1; $n <= $num_digits; $n++) {
$sum = 0; $mult = 2;
for ($i = (strlen($input) - 1); $i >= 0; $i--) {
$sum += ($mult * (int)$input[$i]);
if ((++$mult) > $limit) $mult = 2;
}
if ($x10) {
$digit = (($sum * 10) % 11) % 10;
} else {
$digit = $sum % 11;
if ($digit == 10) $digit = 'x';
}
$input += ($digit);
}
return substr($input, strlen($input) - $num_digits, $num_digits);
}
I wrote an echo calcDigitMod11(200300, 1, 9, true); but it returns the digit to me.
0
I didn't find where the mistaken point x is, I don't know if I'm running away from logic! Here is a table of the JavaScript function of the true digits according to the input value:
calcDigitMod11(200300, 1, 9, true); is return 7
calcDigitMod11(200301, 1, 9, true); is return 5
calcDigitMod11(200302, 1, 9, true); is return 3
calcDigitMod11(200303, 1, 9, true); is return 1

Answer 1

In PHP, you can retrieve characters using index values only for strings. Example:
$str = "98765";
echo $str[1]; // will return 8
If we do the same for integers it will throw a warning.
$num = 98765;
echo $num[1];
Warning: Trying to access array offset on value of type int
To enable errors/warnings in your script, add the below code at the top of your php file
error_reporting(E_ALL);
ini_set('display_errors', '1');
Now, to fix your algorithm, we can use str_split to convert the integer values into an array before using index to extract digits.
<?php
function calcDigitMod11($input, $num_digits, $limit, $x10) {
$mult; $sum; $i; $n; $digit;
if (!$x10) $num_digits = 1;
for ($n = 1; $n <= $num_digits; $n++) {
$sum = 0; $mult = 2;
$input_arr = str_split($input); // Convert string to an array
for ($i = (strlen($input) - 1); $i >= 0; $i--) {
$sum += ($mult * $input_arr[$i]); // Use array to fetch digits
if ((++$mult) > $limit) $mult = 2;
}
if ($x10) {
$digit = (($sum * 10) % 11) % 10;
} else {
$digit = $sum % 11;
if ($digit == 10) $digit = 'x';
}
$input += ($digit);
}
echo "<br/><br/>True Digit=$digit<br/>";
return substr($input, strlen($input) - $num_digits, $num_digits);
}
echo "Output=" . calcDigitMod11(200300, 1, 9, true);
echo "Output=" . calcDigitMod11(200301, 1, 9, true);
echo "Output=" . calcDigitMod11(200302, 1, 9, true);
echo "Output=" . calcDigitMod11(200303, 1, 9, true);
?>
Output:
True Digit=7
Output=7
True Digit=5
Output=6
True Digit=3
Output=5
True Digit=1
Output=4
Working Demo