note 113387 added to mysqli.prepare

If your IDE isn't recognizing $stmt as an object of type mysqli_stmt when you use the traditional perpare: $stmt = mysqli_prepare($link, $query); The following works and is IDE friendly: $stmt = new mysqli_stmt($link, $query); ---- Server IP: 69.147.83.199 Probable Submitter: 208.65.149.19 ---- Manual Page -- http://php.net/manual/en/mysqli.prepare.php Edit -- https://master.php.net/note/edit/113387 Del: integrated -- https://master.php.net/note/delete/113387/integrated Del: useless -- https://master.php.net/note/delete/113387/useless Del: bad code -- https://master.php.net/note/delete/113387/bad+code Del: spam -- https://master.php.net/note/delete/113387/spam Del: non-english -- https://master.php.net/note/delete/113387/non-english Del: in docs -- https://master.php.net/note/delete/113387/in+docs Del: other reasons-- https://master.php.net/note/delete/113387 Reject -- https://master.php.net/note/reject/113387 Search -- https://master.php.net/manage/user-notes.php