PHP: How to call a function with Parameters and accept return value from another URL/Website

Posted on 2016-12-01
Last Modified: 2016-12-02
I have a two websites both running my PHP modules.
Website-1 requires to make a call to a particular PHP function on website-2 with few parameters.
How do I do this in the simplest manner.

Website-1 uses
PHP Version 5.2.17
Question by:Allan_Fernandes
LVL 83

Expert Comment

by:Dave Baldwin
ID: 41908435
You can only 'call' a PHP file on the second host.  There are two methods, GET and POST.  GET is a normal link like you see in your address bar.  It can have the parameters in the query string that is at the end of it.  POST is usually generated by an HTML form.  The parameters are in the form fields.

How are you 'calling' the PHP page on the second site?  You would normally have to duplicate that on the first site for it to work properly.
LVL 54

Expert Comment

by:Julian Hansen
ID: 41908483
Is this

- Server to Server
- Client to server

Option 1
Your PHP script on Server1 needs to call a PHP service on another Server. The exchange is done between the servers without involving the client (Browser)

Option 2
The HTML page POSTS to the server either using a standard Form post or by means of an AJAX (CORS) request

Which one you use depends on the following

- Where do you need the data
- Do you need to stay on the same page or will you transition between pages
- When you "call" to server 2 - is this to get data or to interact with the server as if you had browsed to it.

There is not enough information in your question to give a definitive answer - can you elaborate.

Author Comment

ID: 41908489
I had checked the syntax of _Get and what it seems like it can call only within that PHP module.
like I am already using

echo $_GET["job"]($_GET["args"]);

But if I want to use it thus it is not working

echo $_GET["http://nnn.nnn.nnn/Regpath/Register.php?job=ChkOnlineRegistration&args=".$wRecvText]

I cannot duplicate the PHP module on WebSite-1 because it is being used by my Web Development team.
Whereas Website-2 is a VPS handled only by me. It consists of my critical code.
I want the liberty to move Website-2 as and when I want therefore Website-1 is the stable front for my application's communications.
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Author Comment

ID: 41908496
Just saw your message Hansen. I will elaborate little further. I have a desktop application built in Delphi (not browser based).
I am in the process of giving my users certain online facilities. From Delphi I simply send the required URL in a httpclient and
I get the results back.
LVL 54

Accepted Solution

Julian Hansen earned 500 total points
ID: 41908570
echo $_GET["job"]($_GET["args"]);

Open in new window

The above scares me - you might want to hang a big red sign on your server that says "HACK ME"

What you are doing here is dynamically calling a function based on parameters from a URL

What you want to do rather is something like this

// Make sure you got a job and assign if not
$job = isset($_GET['job']) ? $_GET['job'] : false;

// Now check if there was something in it
if ($job) {
   // obfuscate the name so you can't have people sending arbitary function names in
   $action = "__{$job}";
   // make sure there is such a function
   if (function_exists($action)) {
       $args = isset($_GET['args']) ? $_GET['args'] : null;

       // safely call your function

// call with ?job=doSomethingUseful
function __doSomethingUseful()

Open in new window

That sorts that out.

Now you want to call the function on another server .

To do this we need to use cUrl (
Sample from here
// Because you are using GET we only need the URL
function callRemoteFunction($url)
    // create curl resource
        $ch = curl_init();

        // set url
        curl_setopt($ch, CURLOPT_URL, $url);

        //return the transfer as a string
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);

        // $output contains the output string
        $output = curl_exec($ch);

        // close curl resource to free up system resources

        return $output;
$job = 'ChkOnlineRegistration';
$args = $wRecvText;

$path = URL . "?job={$job}&args={$args}";
$result = callRemoteFunction($path);
// do something with result

Open in new window

Sample available if needed
LVL 109

Expert Comment

by:Ray Paseur
ID: 41908578
What you want is called an Application Programming Interface ("API").  This article discusses the issues and gives code examples.

PHP 5.2 is dangerously out of date, and no longer supported, not even for security issues.  Current is PHP7+, or any version listed under "Download" on the home page of  Time to upgrade!

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