Php count list of digits

Question

I am getting o/p like "11111" and I want to sum all these digits that should become 5. But if I use count count it is showing one only i.e, 1.Rather it should show 5.

Below is my code,

$count = count($inventory['product_id']);

$product_total = $count;
echo $product_total;//o/p => 1.

I need echo $product_total;//o/p => 5.

Answer 1

You can use the following using str_split to get an array with all characters (in your case digits) and using array_sum to get the sum of all the digits:

$digits = "11112";
$arrDigits = str_split($digits);

echo array_sum($arrDigits); //6 (1 + 1 + 1 + 1 + 2)

Demo: https://ideone.com/tZwi9J

Answer 2

Count is used for counting array elements.

What you can do in PHP, is to iterate over a string using either a foreach (not 100% sure) or for loop for this and accessing the elements like array elements by their index:

$str = '111111123545';
$sum = 0;
for ($i = 0; $i < strlen($str); $i++) {
    $sum += intval($str[$i]);
}

print $sum; // prints 26

---

Alternativly, you can split the string using no delimiter and using the array_sum() function on it:

$str = '111111123545';
$sum = array_sum(str_split($str));
print $sum; // prints 26

Answer 3

array_sum(str_split($number));

Answer 4

Another possible way to count the list of digits in PHP is:

// match only digits, returns counts
echo preg_match_all( "/[0-9]/", $str, $match ); 

// sum of digits
echo array_sum($match[0]);

Example:

$ php -r '$str="s12345abas"; echo "Count :".preg_match_all( "/[0-9]/", $str, $match ).PHP_EOL; echo "Sum :".array_sum($match[0]).PHP_EOL;'
Count :5
Sum :15