Get PHP script result from a page with HTML content by AJAX - php

Question

I have a PHP page with HTML content in it. Now I run some PHP codes between HTMLs and I get some results. The fact is, when I try to get a respond from this page by AJAX it'll show me the whole page content and plus the result I was looking for. What can I do to prevent the page from writing extra content. I know whenever something get printed on page it'll go as respond to the AJAX call but I want a way to somehow fix this.
My page file (name: page.php):
<?php echo $Content->GetData('HEADER'); ?>
<div id="Content">
<div id="Page">
<?php if($Content->GetData('PAGE','IS_TRUE')) : ?>
<?php if(NULL !== $Content->GetData('PAGE','TITLE')) : ?>
<?php echo $Content->GetPlugins("Page:" . $Content->GetData('PAGE','ID') . ",BeforeTitle"); ?>
<div id="Title" dir="rtl">
<?php echo $Content->GetData('PAGE','TITLE'); ?>
</div>
<?php echo $Content->GetPlugins("Page:" . $Content->GetData('PAGE','ID') . ",AfterTitle"); ?>
<?php endif; ?>
<div id="Content" dir="rtl">
<div style="float: right; width: 966px; padding: 6px">
<?php echo $Content->GetPlugins("Page:" . $Content->GetData('PAGE','ID') . ",BeforeContent"); ?>
<?php echo $Content->GetData('PAGE','CONTENT'); ?>
<?php echo $Content->GetPlugins("Page:" . $Content->GetData('PAGE','ID') . ",AfterContent"); ?>
</div>
</div>
<?php else : ?>
<div id="Content" dir="rtl">
<div style="float: right; width: 966px; padding: 6px">
There is no page like this in our archives.
</div>
</div>
<?php endif; ?>
</div>
</div>
<?php echo $Content->GetData('FOOTER'); ?>
If I write an address in my browser like this localhost/cload/blog?action=rate it'll go through my redirection list and show the page.php with blog plugin loaded. The problem is I want to call my blog plugin by AJAX through this address but it will first render the page data.
Sorry if this is messy.

Answer 1

I'd suggest modifying page.php to be some functions, primarily a data processing function, and then an output function. Then, when you load the page, put a check in for whether it's an AJAX request, and if so, echo the data you want as JSON, otherwise render the page using the output function.
Alternatively, you could create a second, separate page, but that could be more difficult to maintain than a single file.

Answer 2

Yes, you should check whether your request is a ajax request, if it is so you should change your response to return only the result whatever want.
This php code will give an rough idea on this.
if($request->isXmlHttpRequest()){
return new Response(json_encode($data_array));
}
Hope this will help.