PHP Preg_Match to Find Multiple Occurrences

PHP preg_match to find multiple occurrences

You want to use preg_match_all(). Here is how it would look in your code. The actual function returns the count of items found, but the $matches array will hold the results:

<?php
$string = "/brown fox jumped [0-9]/";

$paragraph = "The brown fox jumped 1 time over the fence. The green fox did not. Then the brown fox jumped 2 times over the fence";

if (preg_match_all($string, $paragraph, $matches)) {
  echo count($matches[0]) . " matches found";
}else {
  echo "match NOT found";
}
?>

Will output:

2 matches found

Php preg_match multiple occurrences, return unique array

The solution using preg_match_all and array_unique functions:

preg_match_all("~\*\*([^/|*]+)(?=[/|])~", $string, $matches);
$result = array_unique($matches[1]);
print_r($result);

The output:

Array
(
    [0] => css
    [1] => php
    [2] => html
)

(?=[/|]) - positive lookahead assertion which matches word that is followed by one of the characters /|

---

Update: to ignore tags from match update regex pattern with the following ~\*\*([^/|*<>]+)(?=[/|])~

Regex php preg_match multiple occurrences in string

Thsi regex should work for you:

<?php 
$ptn = "#(?:By([A-Za-z]+?))(?=By|$)#";
$str = "findByByteByHouseNumber";
preg_match_all($ptn, $str, $matches, PREG_PATTERN_ORDER);
print_r($matches);
?>

this will be the output:

Array
(
    [0] => Array
        (
            [0] => ByByte
            [1] => ByHouseNumber
        )

    [1] => Array
        (
            [0] => Byte
            [1] => HouseNumber
        )

)

Preg_match_all split multiple occurrences

Try this:

$string="59|https://site59.com20|https://site20.com30|https://site30.com16|https://site15.com66|https://site66.com29|https://site29.com";
preg_match_all("/(?:[0-9][0-9](?:\|)(?:https\:\/\/)(.*?)(?=[\d][\d]\||$))|([\d][\d]\|.*)/", $string, $matches);

Results array in $matches:

[0] => 59|https://site59.com
[1] => 20|https://site20.com
[2] => 30|https://site30.com
[3] => 16|https://site15.com
[4] => 66|https://site66.com
[5] => 29|https://site29.com

php find multiple occurence of string in string

Using a regular expression instead of strpos() is going to be you best bet. I've quickly put the following together which works with your example;

\{PID\s=([0-9]*)\}

You can see a working version here

Use of this in PHP would look like;

$re = '/\{PID\s=([0-9]*)\}/';
$str = 'some text {PID =340}  {PID =357}';

preg_match_all($re, $str, $matches);

// Print the entire match result
print_r($matches);

Edit: Edited to return only the actual ID in the matched string. IMO - this is a better solution than the other 2 answers posted as it returns ID's of any length, and only returns ID's matched in the format you've provided.

I've also updated my working example.

Match multiple occurrences in a single regex

You need two entry points, the first is the sentence "There are currently..." until the opening <p> tag, and the second starts at the end of the last match after the <br> tag and the \n newline.

The first result will use the first entry point, the next results will use the second entry point.

\G is the anchor that matches the position at the end of the precedent match. This feature is interesting since the preg_match_all retries to match the pattern until the end of the string. But since \G is initialized with the start of the string at the begining, we need to avoid this case adding (?!\A) (not at the start of the string).

Instead of using .+, I use [^<]+ to avoid to get out of the tag.

To be more readable I use the verbose mode (x modifier) that allows to ignore spaces and to put comments in the pattern. When I need to write literal spaces I put them between \Q and \E. All characters between \Q and \E are seen as literals (except the pattern delimiter) and spaces are preserved.

$pattern = <<<'EOD'
~                    # using this delimiter instead of / avoids to escape all
                     # the slashes

(?:
    # first entry point
    \QThere are currently \E
    [^<]+?
    \Q entries in \E
    [^<]+ </b> </p> \n <p>
  |
    # second entry point
    (?!\A)\G
    <br>\n
)
\K           # removes all that have been matched before from match result
[^<]+        # the string you want
~x
EOD;

if (preg_match_all($pattern, $text, $matches))
    var_dump($matches[0]);

PHP: preg_match_all() - how to find all occurrences of OR seperated substrings with a regex correctly?

The order of the regex is important. I'm not sure if this fully solves the issue the method of doing it this way may be fundamentally flawed but you can try this:

$regex = [];

for($i=0;$i<10;$i++) {
    $str = "";
    for($a=0;$a<10;$a++) {
        if($a > $i) {
            $str .= $a;
            if(strlen($str)>1) {
              $regex[] = $str;
            }
        }
    }
}

usort($regex, function($a,$b){
    return strlen($b) <=> strlen($a);
});

$myregex = '/'.implode('|', $regex).'/';

What it does is make the number sequences an array, then it sorts them by length and orders them the longest sequences first. The end result is this (after matching)

array(1) {
  [0]=>
  array(9) {
    [0]=>
    string(3) "234"
    [1]=>
    string(2) "12"
    [2]=>
    string(4) "6789"
    [3]=>
    string(2) "12"
    [4]=>
    string(3) "123"
    [5]=>
    string(5) "45678"
    [6]=>
    string(2) "12"
    [7]=>
    string(2) "12"
    [8]=>
    string(7) "2345678"
  }
}

Also note the spaceship operator <=> only works in PHP7+

Hope it helps.

Sandbox

and not go to the next chars after a match

I don't think this is possible with regex, if you mean you want to find 23 234 2345 all at once in 2345607 for example. However if it matches a long sequence it only stands to reason that it must match a shorter one, logically. So you could just trim off the right hand number until the length is 2 and get the matches.

---

Related Topics