note 123239 added to function.defined

Dont forget to put the name of your constant into single quotation mark. You will not get an error or a warning. <?php define("AMOUNT_OF_APPLES", 12); if(defined(AMOUNT_OF_APPLES)){ //you won't get an output here echo AMOUNT_OF_APPLES; } ?> so do instead <?php define("AMOUNT_OF_APPLES", 12); if(defined("AMOUNT_OF_APPLES")){ //here you go echo AMOUNT_OF_APPLES; } //output: 12 ?> It took me half an day to see it... ---- Server IP: 208.43.231.9 Probable Submitter: 130.180.65.126 ---- Manual Page -- http://php.net/manual/en/function.defined.php Edit -- https://master.php.net/note/edit/123239 Del: integrated -- https://master.php.net/note/delete/123239/integrated Del: useless -- https://master.php.net/note/delete/123239/useless Del: bad code -- https://master.php.net/note/delete/123239/bad+code Del: spam -- https://master.php.net/note/delete/123239/spam Del: non-english -- https://master.php.net/note/delete/123239/non-english Del: in docs -- https://master.php.net/note/delete/123239/in+docs Del: other reasons-- https://master.php.net/note/delete/123239 Reject -- https://master.php.net/note/reject/123239 Search -- https://master.php.net/manage/user-notes.php