Get parsed PHP file - php

Question

I'm trying to get the contents of a PHP file after it is parsed, and then store it in a variable.
I couldn't get any useful information via Google, except for this one example:
ob_start();
include $file;
$content = ob_get_clean();
But this returns the contents as plain text, i.e.: The <?php and ?> tags are still there, and all code between the tags isn't parsed.
So I wanted to know, how can I do this properly?
update:
This is content of the file which is being included:
Testcontent
<?php echo 'This should be parsed, right?'; ?>

Answer 1

I was using this function several years ago for a sort of a template engine, it seems to do what you need - pass a string with some PHP code inside and it will return it with PHP executed. Surprisingly, it still works :-)
function process_php( $str )
{
$php_start = 0;
$tag = '<?php';
$endtag = '?>';
while(is_long($php_start = strpos($str, $tag, $php_start)))
{
$start_pos = $php_start + strlen($tag);
$end_pos = strpos($str, $endtag, $start_pos); //the 3rd param is to start searching from the starting tag - not to mix the ending tag of the 1st block if we want for the 2nd
if (!$end_pos) { echo "template: php code has no ending tag!", exit; }
$php_end = $end_pos + strlen($endtag);
$php_code = substr($str, $start_pos, $end_pos - $start_pos);
if( strtolower(substr($php_code, 0, 3)) == 'php' )
$php_code = substr($php_code, 3);
// before php code
$part1 = substr($str, 0, $php_start);
// here set the php output
ob_start();
eval($php_code);
$output = ob_get_contents();
ob_end_clean();
// after php code
$part2 = substr($str, $php_end, strlen($str));
$str = $part1 . $output . $part2;
}
return $str;
}

Answer 2

Based on Tyil's last comment he wants to entirety of the php file within a variable:
Tyil, what if my include file looks like this: <?php echo 'test stuff'; $foo = 'one';. What does $content contain and what happens if I try to access $foo from the including file?
#MikeB $content contains the string <?php echo 'test stuff'; $foo = 'one';. var_dump($foo); in the including file returns NULL.
<?php
$file = 'include.php';
$content = file_get_contents($file);
var_dump($content); // (string) "<?php echo 'This should be parsed, right?'; $foo = 'one'; ?>"
include.php:
<?php echo 'This should be parsed, right?'; $foo = 'one'; ?>

Answer 3

If you can modify your $file, than use return in it.
Otherwise cURL it (opens a web page like browser does).

Answer 4

This should definitely work. And it does for me:
[ghoti#pc ~/tmp]$ cat file1.php
#!/usr/local/bin/php
<?php
$file="file2.php";
include($file);
[ghoti#pc ~/tmp]$ cat file2.php
<?php echo "This should be parsed, right?\n"; ?>
[ghoti#pc ~/tmp]$ ./file1.php
This should be parsed, right?
[ghoti#pc ~/tmp]$
You might want to look at your included file (named in $file) and see if there is perhaps some strange character after the initial <?php that might cause it not to be interpreted as a PHP script.
To see a hex dump of what's in the file (so you can see what character actually follows <?php rather than what's displayed in your editor), use: od -c filename.php | less.

Answer 5

To do this, you have to use cURL.
Well, if you want to do it effectively, anyhow.
I know I'm several years late, but I was also looking for a solution to this issue, and I'd like to share the solution. Keep in mind, while this script does indeed get a PHP HTML page parsed, downloading via the web protocol is very slow, at least in my tests.
function GetHTMLPage($url)
{
$ch = curl_init();
$timeout = 5;
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, $timeout);
$data = curl_exec($ch);
curl_close($ch);
return $data;
}
...don't thank me. Thank David Walsh. (https://davidwalsh.name/curl-download)