PHP error: "Cannot pass parameter 2 by reference"

Question

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?

Answer 1

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

Answer 2

First,you shouldn't use DATE_FORMAT when you want to compare date because DATE_FORMAT changes it to string not date anymore,

UPDATE Session 
SET Active = ? 
WHERE SessionDate <= ?

Second, store the value first on a variable and pass it on the paramater

$createDate = mktime(0,0,0,09,05,date("Y"));
$selectedDate =  date('d-m-Y', ($createDate));
$active = 0;
$sql = "UPDATE Session SET Active = ? WHERE SessionDate <= ?";                                         
$update = $mysqli->prepare($sql);
$update->bind_param("is", $active, $selectedDate);  
$update->execute();