php - Run code if not running from an include -


page1.php:

include 'page2.php'; echo $info;

page2.php:

$info = "some info"; if(!included){   echo 'why on page?'; }

the goal of if(!included){ determine if page2.php loaded via include in page1.php, or if user requested page directly. there no function in docs returns such value. how can create such function?

you have 2 options:

a) check debug_backtrace if you're being included

$bt = end(debug_backtrace()); // supposing don't include in functions / other included files etc... if (!empty($bt) && in_array($bt["function"], array("include", "include_once", "require", "require_once"))) {     // you're being included! }

b) define constant, variable etc. in page1.php , check in page2.php if exists (defined or isset)


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