PHP错误：“无法通过引用传递参数2”

Question

我只需要这个不太了解的PHP错误的帮助：

致命错误：在第13行的/web/stud/openup/inactivatesession.php中无法通过引用传递参数2

<?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();

?>

这个错误是什么意思？如何解决此错误？

Answer 1

首先，DATE_FORMAT当您要比较日期时不应该使用，因为DATE_FORMAT将其更改为字符串不再是日期，UPDATE Session&nbsp;SET Active = ?&nbsp;WHERE SessionDate <= ?第二，首先将值存储在变量中，然后将其传递给参数$createDate = mktime(0,0,0,09,05,date("Y"));$selectedDate =&nbsp; date('d-m-Y', ($createDate));$active = 0;$sql = "UPDATE Session SET Active = ? WHERE SessionDate <= ?";&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;$update = $mysqli->prepare($sql);$update->bind_param("is", $active, $selectedDate);&nbsp;&nbsp;$update->execute();