how to pass a variable through the require() or include() function of php?
<p>when I use this:</p>

<pre><code>require("diggstyle_code.php?page=$page_no");
</code></pre>

<p>the warning is :failed to open stream: No error in C:\xampp\htdocs\4ajax\gallery_core.php on line 198</p>

<p>and the error is:</p>

<pre><code>Failed opening required 'diggstyle_code.php?page=1' (include_path='.;C:\xampp\php\PEAR') in C:\xampp\htdocs\4ajax\gallery_core.php on line 198
</code></pre>

<p>value of the variable <code>$page_no</code> is collected beforehand.</p>

<p>But if I omit the <code>'?page=$page_no part'</code> from the argument of the require function, then no error or warning is shown.</p>

<p>I need to pass the variable when I use the require() function. 
Help plz.</p>

<p>Istiaque Ahmed</p>
