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

If you want to run it you could call it with curl and then use the output. — Ryan Matthews
– Ryan Matthews, Commented May 3, 2011 at 19:18
I need this feature to send the HTML to a web service that will generate a PDF. Thanks for the question. — AndyClaw
– AndyClaw, Commented Mar 27, 2013 at 14:03

Gabriel Nahmias · Accepted Answer · 2013-09-16 16:54:06Z

12

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;
}

edited Sep 16, 2013 at 16:54

Gabriel Nahmias

9963 gold badges15 silver badges20 bronze badges

answered May 3, 2011 at 19:28

Robik

6,1574 gold badges33 silver badges41 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Matt Over a year ago

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

Dutchie432 · Accepted Answer · 2011-05-03 19:20:11Z

2

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");

answered May 3, 2011 at 19:20

Dutchie432

29.3k20 gold badges94 silver badges110 bronze badges

Comments

Chris Eberle · Accepted Answer · 2011-05-03 19:20:40Z

2

This should do the trick:

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

answered May 3, 2011 at 19:20

Chris Eberle

48.9k12 gold badges85 silver badges123 bronze badges

Comments

webbiedave · Accepted Answer · 2011-05-03 19:25:11Z

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

edited May 3, 2011 at 19:25

answered May 3, 2011 at 19:20

webbiedave

49k8 gold badges92 silver badges104 bronze badges

1 Comment

Raul Chiarella Over a year ago

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...

jsgoupil · Accepted Answer · 2011-05-03 19:18:31Z

1

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();

answered May 3, 2011 at 19:18

jsgoupil

4,0473 gold badges41 silver badges60 bronze badges

Comments

Michael · Accepted Answer · 2011-05-03 19:25:21Z

1

Output buffering will hold outputted text in memory instead of actually outputting it to the response.

First, create a buffer by calling the ob_start() function. When you are done storing output in the buffer, call ob_get_clean() to return the buffer contents and clear the buffer.

ob_start();
include "../photos_page.php";
$html = ob_get_clean();

answered May 3, 2011 at 19:25

Michael

35.5k17 gold badges79 silver badges112 bronze badges

Comments

akashivskyy · Accepted Answer · 2011-05-03 19:42:50Z

1

You should use file_get_contents("http://yourdomain.com/path/to/photos_page.php") for that.

BUT: If I were you, I would do it in this way:

photos_page.php

<?php

function get_photos_html() {
    $html = // generate html
    return $html;
}

?>

main_file.php

<?php

include('../photos_page.php');

$html = get_photos_html();

?>

edited May 3, 2011 at 19:42

answered May 3, 2011 at 19:18

akashivskyy

45.4k16 gold badges108 silver badges117 bronze badges

5 Comments

user395760 Over a year ago

Nope, he wants the script's output.

akashivskyy Over a year ago

But file_get_contents returns output!

Dutchie432 Over a year ago

@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 Over a year ago

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

Matt Over a year ago

@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.

Lukas Knuth · Accepted Answer · 2011-05-03 19:17:54Z

0

I guess what you posted is pseudo code.

A code you wrote before and which didn't work out well shouldn't be used again. I would rewrite the parts you need and create some functions or even a class which you can use to get whatever you need.

answered May 3, 2011 at 19:17

Lukas Knuth

25.8k16 gold badges87 silver badges114 bronze badges

Comments

jrn.ak · Accepted Answer · 2011-05-03 19:18:52Z

0

If your photos_page.php works like this:

<?php
    // very basic code
    $img = "/path/to/my/image.jpg";
    echo '<img src="' . $img . '">' . PHP_EOL;
?>

Then you could use:

$html = file_get_contents('http://mysite.com/photos_page.php');

But really, you should re-write the code so you don't have to do it this way.

answered May 3, 2011 at 19:18

jrn.ak

36.8k7 gold badges37 silver badges45 bronze badges

Comments

rayalois22 · Accepted Answer · 2020-02-19 19:52:35Z

0

You could also do this if $path returns a value:

$path = '/path/to/your/script.php';
$html = return include $path;

answered Feb 19, 2020 at 19:52

rayalois22

1613 silver badges7 bronze badges

Collectives™ on Stack Overflow

assign output of execution of PHP script to a variable?

10 Answers 10

1 Comment

Comments

Comments

1 Comment

Comments

Comments

5 Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

10 Answers 10

1 Comment

Comments

Comments

1 Comment

Comments

Comments

5 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related