PHP Find all occurrences of a substring in a string - php

Question

I need to parse an HTML document and to find all occurrences of string asdf in it.
I currently have the HTML loaded into a string variable. I would just like the character position so I can loop through the list to return some data after the string.
The strpos function only returns the first occurrence. How about returning all of them?

Answer 1

Without using regex, something like this should work for returning the string positions:
$html = "dddasdfdddasdffff";
$needle = "asdf";
$lastPos = 0;
$positions = array();
while (($lastPos = strpos($html, $needle, $lastPos))!== false) {
$positions[] = $lastPos;
$lastPos = $lastPos + strlen($needle);
}
// Displays 3 and 10
foreach ($positions as $value) {
echo $value ."<br />";
}

Answer 2

You can call the strpos function repeatedly until a match is not found. You must specify the offset parameter.
Note: in the following example, the search continues from the next character instead of from the end of previous match. According to this function, aaaa contains three occurrences of the substring aa, not two.
function strpos_all($haystack, $needle) {
$offset = 0;
$allpos = array();
while (($pos = strpos($haystack, $needle, $offset)) !== FALSE) {
$offset = $pos + 1;
$allpos[] = $pos;
}
return $allpos;
}
print_r(strpos_all("aaa bbb aaa bbb aaa bbb", "aa"));
Output:
Array
(
[0] => 0
[1] => 1
[2] => 8
[3] => 9
[4] => 16
[5] => 17
)

Answer 3

Its better to use substr_count . Check out on php.net

Answer 4

function getocurence($chaine,$rechercher)
{
$lastPos = 0;
$positions = array();
while (($lastPos = strpos($chaine, $rechercher, $lastPos))!== false)
{
$positions[] = $lastPos;
$lastPos = $lastPos + strlen($rechercher);
}
return $positions;
}

Answer 5

This can be done using strpos() function. The following code is implemented using for loop. This code is quite simple and pretty straight forward.
<?php
$str_test = "Hello World! welcome to php";
$count = 0;
$find = "o";
$positions = array();
for($i = 0; $i<strlen($str_test); $i++)
{
$pos = strpos($str_test, $find, $count);
if($pos == $count){
$positions[] = $pos;
}
$count++;
}
foreach ($positions as $value) {
echo '<br/>' . $value . "<br />";
}
?>

Answer 6

Use preg_match_all to find all occurrences.
preg_match_all('/(\$[a-z]+)/i', $str, $matches);
For further reference check this link.

Answer 7

Salman A has a good answer, but remember to make your code multibyte-safe. To get correct positions with UTF-8, use mb_strpos instead of strpos:
function strpos_all($haystack, $needle) {
$offset = 0;
$allpos = array();
while (($pos = mb_strpos($haystack, $needle, $offset)) !== FALSE) {
$offset = $pos + 1;
$allpos[] = $pos;
}
return $allpos;
}
print_r(strpos_all("aaa bbb aaa bbb aaa bbb", "aa"));

Answer 8

Another solution is to use explode():
public static function allSubStrPos($str, $del)
{
$searchArray = explode($del, $str);
unset($searchArray[count($searchArray) - 1]);
$positionsArray = [];
$index = 0;
foreach ($searchArray as $i => $s) {
array_push($positionsArray, strlen($s) + $index);
$index += strlen($s) + strlen($del);
}
return $positionsArray;
}

Answer 9

Simple strpos_all() function.
function strpos_all($haystack, $needle_regex)
{
preg_match_all('/' . $needle_regex . '/', $haystack, $matches, PREG_OFFSET_CAPTURE);
return array_map(function ($v) {
return $v[1];
}, $matches[0]);
}
Usage:
Simple string as needle.
$html = "dddasdfdddasdffff";
$needle = "asdf";
$all_positions = strpos_all($html, $needle);
var_dump($all_positions);
Output:
array(2) {
[0]=>
int(3)
[1]=>
int(10)
}
Or with regex as needle.
$html = "dddasdfdddasdffff";
$needle = "[d]{3}";
$all_positions = strpos_all($html, $needle);
var_dump($all_positions);
Output:
array(2) {
[0]=>
int(0)
[1]=>
int(7)
}

Answer 10

<?php
$mainString = "dddjmnpfdddjmnpffff";
$needle = "jmnp";
$lastPos = 0;
$positions = array();
while (($lastPos = strpos($html, $needle, $lastPos))!== false) {
$positions[] = $lastPos;
$lastPos = $lastPos + strlen($needle);
}
// Displays 3 and 10
foreach ($positions as $value) {
echo $value ."<br />";
}
?>