variable argument in include
<p>I'm sure that I use a wrong way but it's my first attempt to use php. The next code doesn't work. The variable <code>$file</code> is defined locally but I don't know how to use a global variable.</p>

<p><strong>articles.php</strong>
In this file I have a list :</p>

<pre><code>&lt;ol&gt;
  &lt;li&gt;&lt;a &lt;?php $file="a.php"; ?&gt; href="article.php" &gt; title 1 &lt;/a&gt;  &lt;/li&gt;
  &lt;li&gt;&lt;a &lt;?php $file="b.php"; ?&gt; href="article.php" &gt; title 2  &lt;/a&gt;  &lt;/li&gt;
&lt;/ol&gt;
</code></pre>

<p><strong>article.php</strong> in this file I want to use this</p>

<pre><code>  &lt;?php   include("articles/$file"); ?&gt;
</code></pre>

<p>How to define correctly <code>$file</code> ?
Is there a better way to load <code>a.php</code> and <code>b.php</code>  ?</p>
