PHP: Equivalent of include using eval

Question

If the code is the same, there appears to be a difference between:

include \'external.php\';

and

eval(\'?>\' . file_get_conten

Answer 1

As noted by @bwoebi in this answer to my question, the eval substitution does not respect the file path context of the included file. As a test case:

Baz.php:

Foo.php:

' . file_get_contents('Baz.php',  FILE_USE_INCLUDE_PATH)) . "\n";
echo (include 'Baz.php') . "\n";

Result of executing php Foo.php:

$ php Foo.php 
/path/to/file/Foo.php(2) : eval()'d code
/path/to/file/Baz.php

I don't know of any way to change the __FILE__ constant and friends at runtime, so I do not think there is any general way to define include in terms of eval.