assign output of execution of PHP script to a variable?

Question

I made a website, I probably didn't do it like I should have, but I was new to PHP at the time. So in order to save me lots of frustration of trying to re-write a script to display photos on my site, I need to run a *.php file, and make the output if it go into a var called "$html". I know it might sound strange, but that's what I need.

From inside index.php, I include photos.php; In photos.php, I need to declare $html with the output of a script called photos_page.php;

For example: $html = parse_my_script("../photos_page.php");

Thank you

Solution 1

Answer: To do that, you can use PHP's Output buffering/control. Here's some simple function that gets script output and returns it:

Code:

Things used: ob_start() ob_get_clean() is_readable()

function getScriptOutput($path, $print = FALSE)
{
    ob_start();

    if( is_readable($path) && $path )
    {
        include $path;
    }
    else
    {
        return FALSE;
    }

    if( $print == FALSE )
        return ob_get_clean();
    else
        echo ob_get_clean();
}

Usage:

$path = '../photos_page.php';
$html = getScriptOutput($path);

if( $html === FALSE)
{
    # Action when fails
}
else
{
    echo $html;
}

Solution 2

You can use output buffering. This will place all output, that would normally be sent to the client, into a buffer which you can then retrieve:

ob_start();
include '../photos_page.php';
$html = ob_get_contents();
ob_end_clean();

If you wish, you can place this functionality into a function to have it work as you described:

function parse_my_script($path)
{
    ob_start();
    include $path;
    $html = ob_get_contents();
    ob_end_clean();
    return $html;
}

This, of course, assumes that your included file doesn't require the use of global variables.

For more information, check out all the output control functions:

http://www.php.net/manual/en/ref.outcontrol.php

Solution 3

This should do the trick:

ob_start();
require('../photos_page.php');
$html = ob_get_contents();
ob_end_clean();

Solution 4

You'll want to try file_get_contents

$html = file_get_contents("http://www.yourwebsite.com/pages/photos_page.php");

//this will not work since it won't run through web server
//$html = file_get_contents("../photos_page.php");

Solution 5

You want to have a look at the method ob_start() and ob_flush(), ob_get_contents(), etc. http://us.php.net/manual/en/ref.outcontrol.php

This will allow you to output data from your PHP file into a specific variable.

You should think about rewriting though of course :)

So basically:

ob_start();
include('yourfile.php');
$html = ob_get_contents();
ob_end_clean();

View more solutions

Comments

[Admin]

Nope, he wants the script's output.

[akashivskyy]

But file_get_contents returns output!

[Dutchie432]

@Kashiv: If I'm not mistaken, this function returns output only if you are fetching the php page through a web server. file_get_contents("../photos_page.php") won't work. In this case, you will be retrieving the file locally. Needs something like file_get_contents("http://www.domain.com/photos_page.php")

[webbiedave]

This unnecessarily creates a web request which uses many more resources than just simply using an output buffer.

[Matt]

Thanks, your code works. however, my code is so not-cool, that I decided to just include() it from the main index.php file.

[Matt]

@delnan is right. I do want the output. However, the whole stupid variable thing was not working right. Maybe because I was laoding "A" [index.html] which loads "B" [photos.php] which loads "C" [photos_page.php], and ... well it's too complicated to explain, I've decided to work it another way.

[Raul Chiarella]

Hello! I am sorry, i didn't understand your question very well... I have the following PHP line: shell_exec("find ../ -name 'MY_FILE.pdf'") - I want to assign this output (Which returns a directory, with the location of the file) as a usable string to a var called $pdf_path ... I tried using your function but coul'nt figure out how to work with it...