Regular expression missed first occurrence of target string - php

Question

I am using regular expression to fetch both text1 and text2 in the following html code. Here is what I am using:
/<div\s?class="right-col">[\s\n\S]*<a[\s\n]?[^>]*>#(.*)<\/a>/
but apparently I missed text1, only got text2(here is the link to my problem).
<div class="right-col">
<h1>
title1
</h1>
<p>some text here</p>
<div class="some-class">
<div class="left">
<span>some text here </span>
</div>
<div class="postmeta"><a href="url-link-here" >#text1</a> </div>
</div>
<div class="right-col">
<h1>
title2
</h1>
<p>some text here</p>
<div class="some-class">
<div class="left">
<span>some text here </span>
</div>
<div class="postmeta"><a href="url-link-here" >#text2</a> </div>
</div>
Can you guys tell me what went wrong in my regular expression? Is there a better way to capture both title1, title2 and text1, text2?

Answer 1

Using a regular expression here is not the best way to do it. It's bad practice. You should be using a DOM/XML parser to do this.
I like using PHP's DOMDocument class. Using XPath, we can quickly find the elements you want
$dom = new DOMDocument;
$dom->loadHTML($html);
$xPath = new DOMXPath($dom);
$aTags = $xPath->query('//div[#class="some-class"]//a[starts-with(text(), "#")]');
foreach($aTags as $a){
echo $a->nodeValue;
}
DEMO: http://codepad.viper-7.com/QHOXzH

Answer 2

This is a fairly common issue with regular expressions as they are greedy. [\s\S]* (the \n is not needed) matches for the first '<' and 'a' and since it's greedy it will match those and continue. Adding a ? makes it not greedy and using your link returns both text1 and text2.
The short answer is to replace [\s\n\S]* with [\s\S]*? but as others have mentioned, this is probably not a good use of regular expressions.