PHP error: "Cannot pass parameter 2 by reference"
mysqli, php
Solution
The error means that the 2nd argument is expected to be a reference to a variable.
Since you are not handing a variable but an integer of value 0, it generates said error.
To circumvent this do:
$a = 0;
$update->bind_param("is", $a, $selectedDate); //LINE 13
In case you want to understand what is happening, as opposed to just fixing your `Fatal error`, read this: http://php.net/manual/en/language.references.pass.php
Problem
I just need help on this PHP error which I do not quite understand: Fatal error: Cannot pass parameter 2 by reference in /web/stud/openup/inactivatesession.php on line 13 ``` <?php error_reporting(E_ALL); include('connect.php'); $createDate = mktime(0,0,0,09,05,date("Y")); $selectedDate = date('d-m-Y', ($createDate)); $sql = "UPDATE Session SET Active = ? WHERE DATE_FORMAT(SessionDate,'%Y-%m-%d' ) <= ?"; $update = $mysqli->prepare($sql); $update->bind_param("is", 0, $selectedDate); //LINE 13 $update->execute(); ?> ``` What does this error mean? How can this error be fixed?