Operation on String - Substrings: Get Positions and Count - php

Question

I have a problem with operations on a string in PHP, I have one string like this:
$words = "Ala ma kota a kot ma ale";
How do I get the number of appearances of al in this long string $words? Additionally, I need the index of the beginning of all appearances of al.
$count = substr_count($words, 'al');
I tried it with the substr_count(), but it only returned the count. I need the index of the appearance as well.
EDIT Adding expected output:
number of al: 2, at index: 0, at index: 22

Answer 1

This is easily accomplished with preg_match_all() using PREG_OFFSET_CAPTURE flag:
$words = "Ala ma kota a kot ma ale. All in Valhalla shall recall the fall.";
preg_match_all('~(al)~i', $words, $matches, PREG_OFFSET_CAPTURE);
print_r($matches[1]);
See a live demo at https://3v4l.org/Frsa5. We have the i modifier for case-insensitive matching. If you want case-sensitive matching, remove it. If you want only als at the start of words, use \bal (\b = word boundary). The result is an array with matches and offsets, as follows:
Array [
[0] => [
[0] => Al
[1] => 0
]
[1] => [
[0] => al
[1] => 21
]
[2] => [
[0] => Al
[1] => 26
]
[3] => [
[0] => al
[1] => 34
]
[4] => [
[0] => al
[1] => 37
]
[5] => [
[0] => al
[1] => 44
]
[6] => [
[0] => al
[1] => 51
]
[7] => [
[0] => al
[1] => 60
]
]
Edit: Since there's nothing else being matched, you don't really need the (al) capture group. You can also just remove the brackets, match ~al~i, and get the results in $matches[0] (containing full pattern matches). I've left it as is with the capture group in place, in case you may want to use more complex matching rules in the future (& being lazy to update the demo).

Answer 2

You could build a loop that continuously searches the array, until you run out of count, probably. Or you could wing it and make a loop that continuously does strpos() and count it in the end. Note that you need to somehow make sure you're not checking the same position over and over, that's what the $position++ is there in this piece of code.
Here, this will output exactly what you asked for in the comment, with the correct position that is. (I didn't see the comment before)
$position = 0;
$words = strtolower("Ala ma kota a kot ma ale");
$needle = 'al';
$positions = [];
do {
$position = strpos($words, $needle, $position);
if ($position !== false) {
array_push($positions, $position);
$position++;
}
} while ($position);
echo 'number of al: '.count($positions);
foreach ($positions as $position) {
echo ', at index: '.$position;
}
Output:
number of al: 2, at index: 0, at index: 21
Edit: As noted by eis, this could be simplified down to
<?php
$position = 0;
$words = strtolower("Ala ma kota a kot ma alet ma ale");
$needle = 'al';
$positions = [];
$position = -1;
while (($position = strpos($words, $needle, $position + 1)) !== false) {
array_push($positions, $position);
}
echo 'number of al: ' . count($positions);
foreach ($positions as $position) {
echo ', at index: ' . $position;
}
Edit: As Markus AO said, you might want to base your position movement (position++;) on the length of the needle. I think that depends on the way you wish to match.
a) If you want to match all appearances (aba → abababa = 3 ... match may occur within the previous match,) keep it this way.
b) If you want to match all 'full' appearances (aba → abababa = 2,) increment the position by the needle's length.