How to return raw php code?

Question

I have this code for example:

ob_start();
echo "hello there";
$output = ob_get_contents();
return $output;

When I run it, I get back:

hello there

But how can I get back

echo "hello there";

Is there a way to do this easily?

I don't quite understand, why is the echo code the only one returned? are you only interested in returning echo calls? — Samuel Allan
– Samuel Allan, Commented Jul 10, 2016 at 9:02
?>echo "hello there";<?php - drop out of PHP mode and anything you type (PHP code or otherwise) shows for the world to see. — Niet the Dark Absol
– Niet the Dark Absol, Commented Jul 10, 2016 at 9:02
seems like You want to get code of some encoded php file by executing between ob_* (: if Yes, so it's not done like this. — num8er
– num8er, Commented Jul 10, 2016 at 9:04
@num8er: if you say it's not done like this - can you elaborate how it is done? — Taapo
– Taapo, Commented Jul 10, 2016 at 9:45
I mean if You want to get source of obfuscated-encoded script, so You've to analyze file manually and do complex variable mapping and etc. I cannot show it's long procedure. — num8er
– num8er, Commented Jul 10, 2016 at 10:41

edhurtig · Accepted Answer · 2016-07-10 20:36:51Z

4

To output arbitrary text as-is you can close the PHP script and then reopen it. Anything between the closing and opening tags is output as-is

ob_start();
?>echo "hello there";
<?php
$output = ob_get_contents();
return $output;

edited Jul 10, 2016 at 20:36

answered Jul 10, 2016 at 9:05

edhurtig

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5 Comments

Denis Matafonov Over a year ago

yes, but this kind of 'script' will be useless - you cant run it

Taapo Over a year ago

It's actually useful if you have a script that needs to generate a hard-coded script ;)

Taapo Over a year ago

Can you conditionally drop in and out of PHP?

edhurtig Over a year ago

Yes you can @Taapo just drop out in the body of an if/else statement

Taapo Over a year ago

@TheH Problem with that approach is that you have to echo twice in an if/else statement, not? If you use <? if ($statement == true) { ?>echo "hello";<? } else { echo "hello"; } ?> -- I would love to prevent the double work. The following won't work, but I'm wondering if there is an other way? <? if ($statement == true) { ?> } echo "hello"; { <? } ?>

Denis Matafonov · Accepted Answer · 2016-07-10 09:40:32Z

2

ob_get_contents will return the echoed output, so you can't use it to show actual code.

To simply print out code, I would try this:

$code = file_get_contents('your_code.php');
echo "<pre>{$code}</pre>";

Also you can write you code separatle as text and echo it or eval (if you need execution).

$string = 'cup';
$name = 'coffee';
$str = 'This is a $string with my $name in it.';
echo $str. "\n";
eval("\$str = \"$str\";");
echo $str. "\n";

result:

This is a $string with my $name in it.
This is a cup with my coffee in it.

edited Jul 10, 2016 at 9:40

answered Jul 10, 2016 at 9:07

Denis Matafonov

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Comments

revo · Accepted Answer · 2016-07-10 09:09:13Z

1

By representing it as a string:

ob_start();
$str = <<<STR
echo "hello there";
STR;
echo $str;
$output = ob_get_contents();
return $output;

answered Jul 10, 2016 at 9:09

revo

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