Requesting information from one php file to another using only php - php

Question

I am pretty new to php so I'm having trouble with POSTing.
I am trying to transfer information between 2 php files where send_var.php sends a command by POST and get_var.php executes some data manipulation and returns the response.
The send_var.php is as follows:
<?php
$url = "./get_var.php";
$fields = array('response' => "323243");
$data = http_build_query($fields);
// echo $data."<br />";
$context = stream_context_create(array(
'http' => array(
'method' => 'POST',
'header' => "Content-type: text/html\r\n",
'content' => $data
),
));
$out = file_get_contents($url, false, $context);
echo "Info from get_var : " . $out;
?>
And the get_var.php is :
<?php
$arg1 = isset($_POST['response']) ? $_POST['response'] : null;
if($arg1 == '')
{
echo "No Command! ";
}
if($arg1 != "")
{
echo $_POST['response'];
}
else
{
$_POST['response'] = "123456";
echo $_POST['response'] . " end of get_var";
}
?>
This code was extracted from other examples on stack overflow. The only output I get is "Info from get_var :"
Obviously I'm missing some pretty fundamental knowledge. If someone can help it would be much appreciated. I'm executing this under XAMPP.

Answer 1

In order to run a PHP script, you have to access it through the webserver. So the URL needs to be an http: URL, not just a filename:
$url = "http://localhost/path/to/get_var.php";
If you just use a filename, file_get_contents() will just return the PHP source code, it won't run it.
Also, your Content-type header is wrong, it should be application/x-www-form-urlencoded, not text/html (that's the content type of the response, your context specifies the type of the POST data).