Pass varibles to include

[belick]

I am tring to pass a POST to an include in some way, here is what I had in mind:

 

$_POST[anything] = 'something';

 

into:

 

include "http://www.domainname.com/file.php";

 

when file is:

 

<?PHP

  echo "$_POST[anything]";

?>

 

Thanks

[cunoodle2]

Maybe you could use Curl..

 

<?php
//set up variables
$url = "http://www.domainname.com/file.php";
$fields = "anything=something";

//execute Curl
$c = curl_init($url);
curl_setopt($c, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($c, CURLOPT_CONNECTTIMEOUT, 0);
curl_setopt($c, CURLOPT_POSTFIELDS, $fields);   //sending over post variables here
$data = curl_exec($c);
curl_close($c);

 

While looking at some of my code I found another example on line.  Here it is..

 

<?php
//curl6.php

// populate the array containing the values to be posted.
$postfields = array();
$postfields['field1'] = urlencode('value1');
$postfields['field2'] = urlencode('value2');

$ch = curl_init();

// Follow any Location headers
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);

curl_setopt($ch, CURLOPT_URL, 'http://www.example.co.za/recipient2.php');
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);

// Alert cURL to the fact that we're doing a POST, and pass the associative array for POSTing.
curl_setopt($ch, CURLOPT_POST, 1);

curl_setopt($ch, CURLOPT_POSTFIELDS, $postfields);

$output = curl_exec($ch);
curl_close($ch);
print $output;
?>

 

Good luck

[cooldude832]

I am tring to pass a POST to an include in some way, here is what I had in mind:

 

$_POST[anything] = 'something';

 

into:

 

include "http://www.domainname.com/file.php";

 

when file is:

 

<?PHP

  echo "$_POST[anything]";

?>

 

Thanks

 

what you are saying is simply impossible because an included/required/required_once file is like pasting its raw unsent source code into it so it has no real location to send.