How to pass parameters with `include`?
<p>I have a PHP script called customers.php that is passed a parameter via the URL, e.g.:</p>

<p><a href="http://www.mysite,com/customers.php?employeeId=1" rel="nofollow">http://www.mysite,com/customers.php?employeeId=1</a></p>

<p>Inside my customers.php script, I pick up the parameter thus:</p>

<pre><code>$employeeId = $_GET['employeeId'];
</code></pre>

<p>That works just fine. I also have an HTML file with a form that inputs the parameter, and runs another php script, viz:</p>

<pre><code>&lt;form method="post" action="listcustomers.php"&gt;
  Employee ID:&amp;nbsp; &lt;input name="employeeId" type="text"&gt;&lt;br&gt;
  &lt;input type="submit" value="Show"&gt;
&lt;/form&gt;
</code></pre>

<p>Then in my listcustomers.php script, I pick up the parameter so:</p>

<pre><code>$employeeId = $_POST['employeeId'];
</code></pre>

<p>So far so good. But his is where I am stuck. I want to run my customers.php script, and pass it the parameter that I pave picked up in the form. I am sure this is really simple, but just cannot get it to work. I have tried:</p>

<pre><code>include "customers.php?employeeId=$employeeId&amp;password=password";
</code></pre>

<p>But that does not work. Nor does anything else that I have tried. Please help me.</p>
