Php parse string error - php

Question

I am extracting files from a string which can be entered by a user or taken from reading a page source.
I want to extract all .jpg image URLs
So, I am using the following (example text shown) but a) it only returns the first one and b) it misses off '.jpg'
$word1='http://';
$word2='.jpg';
$contents = 'uuuuyyyyyhttp://image.jpgandagainhereitishttp://image2.jpgxxxxcccffff';
$between=substr($contents, strpos($contents, $word1), strpos($contents, $word2) - strpos($contents, $word1));
echo $between;
Is there maybe a better way to do this?
In the case of parsing a web page I cannot use a simple DOM e.g. $images = $dom->getElementsByTagName('img'); as sometimes the image references are not in standard tags

Answer 1

You can do something like this :
<?php
$contents = 'uuuuyyyyyhttp://image.jpgandagainhereitishttp://image2.jpgxxxxcccffff';
$matches = array();
preg_match_all('#(http://[^\s]*?\.jpg)#i',$matches);
print_r($matches);

Answer 2

You can either do this using preg_match_all (as previously answered) or alternatively use the following function.
It simply explodes the original string, checks all parts for a valid link and adds it to the array, that's getting returned.
function getJpgLinks($string) {
$return = array();
foreach (explode('.jpg', $string) as $value) {
$position = strrpos($value, 'http://');
if ($position !== false) {
$return[] = substr($value, $position) . '.jpg';
}
}
return $return;
}