PHP function getAllBetweenStr using regular expression

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Get content between two strings PHP

I am trying to create a php function that will return an array of all occurrences of text between two substrings. I am useless at regular expression and would find this very handy, I'm sure there is already a very nice one out there but I can't find it.

    function getAllBetweenStr($input, $start, $end)
    {
         //DO STUFF

         //RETURN ARRAY
    }


$inputStr ="<a href='http://formaco4/en/abc' class='currentLink' hreflang='en'>abc1/a>
<a href='http://formaco4/en/abc/b1'  hreflang='en'>abc2</a>
<a href='http://formaco4/en/abc/b2'  hreflang='en'>abc3</a>
<a href='http://formaco4/en/abc/b3'  hreflang='en'>abc4</a>";


$linkStr = array();
$linkStr = getAllBetweenStr($inputStr, "http://formaco4/en/'", "'  hreflang")

The return array would be like this Array([0] => "abc/b1", [1] => "abc/b2", ... , [n] => "gingers/rule")

I have this function already if it helps:

    function getBetweenStr($string, $start, $end)
    {
        $string = " ".$string;
        $ini = strpos($string,$start);
        if ($ini == 0) return "";
        $ini += strlen($start);    
        $len = strpos($string,$end,$ini) - $ini;
        return substr($string,$ini,$len);
    }

and this is actually what preg_match / preg_match_all (in your case) is made for ... — F. Müller, Oct 05 '12 at 13:25
you define $linkstr as a array. Then you define $linkstr as a string (returning the substr) when you want to have it as an array you have to define it as $link[]. — Jordi Kroon, Oct 05 '12 at 13:27
*(related)* [How to parse and proccess HTML with PHP](http://stackoverflow.com/questions/3577641/how-to-parse-and-process-html-with-php/3577662#3577662) — Gordon, Oct 05 '12 at 14:30

score 1 · Accepted Answer · answered Oct 05 '12 at 13:32

I believe something like this will work - It uses preg_match_all to capture anything between the two $start and $end strings.

function getBetweenStr($string, $start, $end)
{
    preg_match_all( '/' . preg_quote( $start, '/') . '(.*?)' . preg_quote( $end, '/') . '/', $string, $matches);
    return $matches[1];
}

score 0 · Answer 2 · answered Oct 05 '12 at 13:32

Mine doesn't use the preg_match function. Hopefully you will find this useful

Try this, I've also provided a better function for you (: :

function stribet($inputstr, $delimiterLeft, $delimiterRight) {
    $posLeft = stripos($inputstr, $delimiterLeft) + strlen($delimiterLeft);
    $posRight = stripos($inputstr, $delimiterRight, $posLeft);
    return substr($inputstr, $posLeft, $posRight - $posLeft);
}

$links = array("<a href='http://formaco4/en/abc' class='currentLink' hreflang='en'>abc1/a>", "<a href='http://formaco4/en/abc/b1'  hreflang='en'>abc2</a>");

foreach ($links as $linkstr){
$linkstr = stribet($linkstr, "http://formaco4/en/'", "'  hreflang='");
//echo $linkstr;
$data[] = $linkstr;
}

print_r($data);

if you would just like to stick to normal string function, try exploding <a then use the foreach loop to get your data.

Something like

$inputStr ="<a href='http://formaco4/en/abc' class='currentLink' hreflang='en'>abc1/a>
<a href='http://formaco4/en/abc/b1'  hreflang='en'>abc2</a>
<a href='http://formaco4/en/abc/b2'  hreflang='en'>abc3</a>
<a href='http://formaco4/en/abc/b3'  hreflang='en'>abc4</a>";
$inputArray = explode ("<a", $inputStr);
  foreach ($inputArray as $linkstr){
    $linkstr = stribet($linkstr, "http://formaco4/en/'", "'  hreflang='");
    //echo $linkstr;
    $data[] = $linkstr;
    }

    print_r($data);

score 0 · Answer 3 · answered Oct 05 '12 at 13:32

0

It's really worth learning regular expressions. They're very powerful and useful:

preg_match_all('|'.preg_quote($start).'(.*?)'.preg_quote($end).'|', $input, $linkStr);

In your example this would build a regular expression of:

|http:\/\/formaco4\/en\/\(.*?)  hreflang|

The parens () tell it to capture that part of the string. The ".*?" within the parens instruct it to catch all characters up to what follows.

answered Oct 05 '12 at 13:32

Matt S

14,976
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Need to escape `$start` and `$end`, otherwise they can break the regex. – nickb Oct 05 '12 at 13:34
Thanks, I thought of that after I submitted. I was adding it as you typed your comment :) – Matt S Oct 05 '12 at 13:36
Using `|` as delimiter is quite irregular, especially since it's also used for alternation :) – Ja͢ck Oct 05 '12 at 13:53

F. Müller · Answer 4 · 2012-10-05T13:39:40.560

0

Try this one (untested):

function find_occurencies($start, $search, $end)
{
    $start = stripslashes($start);
    $end = stripslashes($end);
    $search = stripslashes($search);

    $pattern = '%'. $start . '([\s\w]+)?' . $end . '%';
    preg_match_all($pattern, $search, $matches);

    // Print the matches (should)
    echo '<pre>';
    print_r($matches);
    echo '</pre>';
}

Hope it works.

edited Oct 05 '12 at 13:39

answered Oct 05 '12 at 13:34

F. Müller

3,969
8
38
49

Need to escape `$start` and `$end`, otherwise they can break the regex. – nickb Oct 05 '12 at 13:34
well, yes ... I was still updating ;) – F. Müller Oct 05 '12 at 13:35
Warning: preg_match_all() [function.preg-match-all]: Compilation failed: nothing to repeat at offset 0 – Tristanisginger Oct 05 '12 at 13:54
It is nearly there but some of the characters in the input strings break the regex – Tristanisginger Oct 05 '12 at 14:00

score 0 · Answer 5 · answered Oct 05 '12 at 13:50

<?php

function getAllBetweenStr($string, $start, $end)
    {
        $array_to_return = array();
        for($i=0; $i<count(explode($start,$string))-1; $i++) {
        $str_0 = explode($start,$string);
        $str_1 = explode($end, $str_0[1]);
        $array_to_return[$i] = $str_1[0];
        }
        return $array_to_return;
    }

$inputStr ="<a href='http://formaco4/en/abc' class='currentLink' hreflang='en'>abc1</a>
<a href='http://formaco4/en/abc/b1'  hreflang='en'>abc2</a>
<a href='http://formaco4/en/abc/b2'  hreflang='en'>abc3</a>
<a href='http://formaco4/en/abc/b3'  hreflang='en'>abc4</a>";
$linkStr = getAllBetweenStr($inputStr, "hreflang='en'>", "</a>");
print_r ($linkStr);
?>

PHP function getAllBetweenStr using regular expression

5 Answers5