get resulting content of a file without including it

Question

01.php

<div class='title'>title</div>

02.php

<?php include('01.php');?>

some function:

$a = file_get_contents('02.php');
echo $a;

Result:

<?php include('01.php');?>

Is there a way to refer 02.php (without including it in the current file) and get this:

<div class='title'>title</div>

This is a simplified example. In reality 02.php is a large file with 01.php included somewhere inside.

Why can't you just use the `include()`? I'm not sure I understand that. Regardless, if that is not an option, you need to look into using the output buffer along with `file_get_contents`. — Jeremy Harris, Dec 20 '18 at 20:31
https://stackoverflow.com/questions/2832010/what-is-output-buffering — Jeremy Harris, Dec 20 '18 at 20:34
Not always an option or a good option, but `$a = file_get_contents('http://example.com/02.php');` — AbraCadaver, Dec 20 '18 at 21:42

score 1 · Accepted Answer · answered Dec 20 '18 at 21:04

If you want to parse the php file and save the result to a variable instead of outputting it you could use:

ob_start();
include("01.php");
$a = ob_get_contents();
ob_end_clean();

You will have the parsed php in $a but it will not be output to the user. More info: http://php.net/manual/en/ref.outcontrol.php

1 Answers1