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    <p>Regular expressions are <a href="http://www.skdevelopment.com/php-regular-expressions.php" rel="nofollow noreferrer">greedy</a> by default (at least PCREs). Usually you could circumvent this:</p> <pre><code>/a+/ # selects the whiole string from "aaaaaaa" /a+?/ # selects only "a" </code></pre> <p>In your case, this doesn't help much, since you can't just throw in a question mark somewhere. The only possibility is to <strong>re-order</strong> your search array and instantly <strong>replace</strong> the found places. Search <strong>first</strong> for <code>s:10</code> and <strong>second</strong> for <code>s:1</code>, and use <a href="http://php.net/preg_replace" rel="nofollow noreferrer">preg_replace()</a> instead of the matching. This way, the second doesn't find the first anymore.</p> <p><strong>Another possibility:</strong> Split your search array in two. If you know, that the one always has the structure 's:' plus digits, you could have your regexp in this second loop like</p> <pre><code>Preg_Match_All ( '/' . Preg_Quote ( $smilie ) . '(?![0-9])/i', $Content, $matches ); </code></pre> <p>with <code>(?![0-9])</code> a <a href="http://de.php.net/manual/en/regexp.reference.assertions.php" rel="nofollow noreferrer">look ahead expression</a> looking for any <em>non</em>-digit.</p> <p><strong>And a third one:</strong> If you allow (== convert) smileys only at certain places, you could use this:</p> <pre><code>Preg_Match_All ( '/\b' . Preg_Quote ( $smilie ) . '\b/i', $Content, $matches ); </code></pre> <p><code>\b</code> is a "word boundary", usually any not-(letter, digit, underscore). Drawback is obviously, that not all smileys (like "abc;-)xyz") will be found.</p>
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    1. COI don't think this will work, because he starts a new regular expression search for each smiley.
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      1. USFortega
    2. COlike Fortega said, this won't work for me. It could if I would replace the found smile as soon as I would find it, but I have to validate first only then convert the text smilies to images if validation passes ...
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      1. USJan Hančič
    3. COBut if the replacement is made by the first regular expression already, then the second regex won't find s:1. +1
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      1. USRob Fonseca-Ensor
 

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