计算字符串中出现的数字数量

I was wondering if there's a combo of functions or a direct function that can count how many numbers appears in a string, without use a long-way as str_split and check every character in a loop.

From a string like:

fdsji2092mds1039m

It returns that there's 8 numbers inside.

You can use filter_var() with the FILTER_SANITIZE_NUMBER_INT constant, then check the length of the new string. The new string will contain only numbers from that string, and all other characters are filtered away.

$string = "j3987snmj3j";
$numbers = filter_var($string , FILTER_SANITIZE_NUMBER_INT);
$length = strlen($numbers); // 5
echo "There are ".$length." numbers in that string";

Note that each number will be counted individually, so 137 would return 3, as would 1m3j7.

Live demo

Other solution:

function countNumbers(string $string) {
    return preg_match_all('/\d/', $string, $m);
}

You can use regular expression

Try like this:

$myString = 'Som3 Charak1ers ar3 N0mberZ h3re ;)';
$countNumbers = strlen((string)filter_var($myString, FILTER_SANITIZE_NUMBER_INT));
echo 'Your input haz ' . $countNumbers . ' digits in it, man';

You can also make a function out of this to return only the number, if you need it.

Following code does what you intend to do:

<?php
$string = 'dsfds98fsdfsdf8sdf908f9dsf809fsd809f8s15d0d';
$splits=str_split($string);
$count=0;
foreach ($splits as $split){
    if(is_numeric($split)){
        $count++;
    }
}

print_r($count);

Output: 17