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    <p><strong>Use <a href="http://www.regular-expressions.info/lookaround.html" rel="nofollow">lookarounds</a></strong></p> <p>Here we're making sure there is a preceding <code>(</code> character, then we look for text we would see in a date formatted like your example. This little bit of code says ALLOW for alpha numeric characters, a literal space character, and a comma, as well as digits <code>([A-Za-z ,\d]+)?</code>. The <code>+</code> character means <em>at least 1</em>. It's not as greedy as <code>.*</code> or <code>.+</code>. I'm surrounding it with parenthesis and then adding a <code>?</code> character to make it not required. It works similar to your <code>|</code> <em>or</em> statement logically because it will still find the year, but we're not making PHP do more work by parsing another check. Then we find the year (always 4 digits <code>{4}</code>). Then we check to make sure it's followed by a literal <code>)</code> character. The look behind <code>(?&lt;=\()</code> and the look ahead <code>(?=\))</code> will find a match, but they are not included in the match results, leaving your answer clean.</p> <p>Since <a href="http://us1.php.net/manual/en/function.preg-match.php" rel="nofollow"><code>preg_match()</code></a> returns an <code>array()</code> we're catching the first element in the array. If you're looking for multiple matches in the same string you can use <code>preg_match_all</code>.</p> <pre><code>$data = '&lt;a href="not important"&gt; &lt;span class="theclass"&gt;data (not important)&lt;/span&gt;&lt;/a&gt; &lt;span class="anotherclass"&gt;extra data (October 1, 2010)&lt;/span&gt; &lt;span class="anotherclass"&gt;extra data (2011)&lt;/span&gt;'; $pattern = '!(?&lt;=\()([A-Za-z ,\d]+)?[\d]{4}(?=\))!'; $res = preg_match_all($pattern,$data,$myDate); print_r($myDate[0]); </code></pre> <p><strong>output</strong></p> <pre><code>Array ( [0] =&gt; October 1, 2010 [1] =&gt; 2011 ) </code></pre> <p>If you're only looking for one match you would change the code to this:</p> <pre><code>$res = preg_match($pattern,$data,$myDate); echo($myDate[0]); </code></pre> <p><strong>Output</strong></p> <pre><code>October 1, 2010 </code></pre> <p>Another way to write the pattern would be like this... we've removed the parenthesis (grouping) and the plus <code>+</code> modifier followed by the conditional <code>?</code>, but left the first set. Then we're using a <code>*</code> to make it conditional. The difference is with preg_match and preg_match_all, any groupings are also stored in the array. Since this isn't a group, then it will not store extra array elements.</p> <pre><code>$pattern = '!(?&lt;=\()[A-Za-z ,\d]*[\d]{4}(?=\))!'; </code></pre>
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    1. COi have added an extra condition that the year is ALWAYS present. and could u please explain ur expression. thanks :)
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    2. COah ok well that works when both the date and year is present, but not only the year :)
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    3. COyour code works perfect when both the date and year is present, but not only the year :) is preg_match able to meet the requirements of my condition :)
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