Php is not working but iframe is working fine

I have few sets of code to display right side of my webpage. I am using this place to show some user information and some short of ads. For ads I am using that code in my database. Now the problem is that I want to display one external php file which is located in the same directory where I am trying to include or require this php file.

But I dont know why this is not working. I tried with iframe that works fine but I want to use php include but it is not working .please tell where I am doing mistake. Here is my code:

if (is_array($data['blocks'])) {
        $output .= '<div id="appside">';
        foreach($data['blocks'] as $block) {
            if (is_array($block)) {
                $output .= '
                <div class="block">';
                if ($block['title']) {
                    $output .= '<div class="block_title">'.$block['title'].'</div>';
                }
                $output .= '<div class="block_content">
                '.$block['content'].
                '</div>
                </div>
                ';
            }
        }
        $output .= get_gvar('theme_block_sidebar').

        '<div><?php include("/home/xxxxxxx/public_html/xxxxxxxa.php"); ?></div>


        </div>

                ';// end of app_sidebar

I tried with all short of alternatives but its printing raw php code. i开发者_运维问答ts not getting executed. I tried to put <?php include(\'/home/xxxxxxx/public_html/xxxxxxxa.php\'); ?> and all other sort of alternatives but its printing raw php code.

---

'<div><?php include("/home/xxxxxxx/public_html/xxxxxxxa.php"); ?></div>

PHP won't execute inside a string (which is all the above is).

You can't really concatenate an include statement (unless the included file returns something via return). Given I can't tell if your code snippet is inside a function or how it's executed, my best solution is to use output buffering.

$output .= get_gvar('theme_block_sidebar') . '<div>';

ob_start();
include "/home/xxxxxxx/public_html/xxxxxxxa.php";
$output .= ob_get_contents();
ob_end_clean();

$output .= '</div>'

---

if the file you wish to include is in the same directory as this file, then you should include the filename only

<?php include('filename.php'); ?>

if it is in a higher level or in a directory or a sub-directory in a higher level, then you should move up to this level then down to the file.

<?php include('../../path/to/file.php'); ?>

where ../../ is going up 2 levels, you can set it to as many levels you want to go up.

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If you want to get the output from an external php file in a variable, you can get it using the PHP buffering functions:

ob_start();
include("/home/xxxxxxx/public_html/xxxxxxxa.php");
$output_from_php_file = ob_get_contents();
ob_end_clean();

Then you can append $output_from_php_file to your $output variable wherever you want.