How to pass parameters with `include`?

Question

I have a PHP script called customers.php that is passed a parameter via the URL, e.g.:

http://www.mysite,com/customers.php?employeeId=1

Inside my customers.php script, I pick up the parameter thus:

$employeeId = $_GET['employeeId'];

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

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

Then in my listcustomers.php script, I pick up the parameter so:

$employeeId = $_POST['employeeId'];

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:

include "customers.php?employeeId=$employeeId&password=password";

But that does not work. Nor does anything else that I have tried. Please help me.

Answer 1

You can't add a query string (Something like ?x=yyy&vv=www) onto an include like that. Fortunately your includes have access to all the variables before them, so if $_GET['employeeId'] is defined before you call:

include 'customers.php';

Then customers.php will also have access to $_GET['employeeId'];