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我只是需要这个PHP错误的帮助,我不太明白:

致命错误:无法在第13行的/web/stud/openup/inactivatesession.php中通过引用传递参数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？如何解决这个错误？

当我运行此代码以更新我的 likeOne 列并将其设为空 ("")....

    $sql11 = $conn->prepare("UPDATE UserData SET likedOne=? WHERE username=?");
    $sql11->bind_param('ss',"",$Username);
    $Username = "netsgets";
    $sql11->execute();

我收到这个错误....

1 Fatal error: Cannot pass parameter 2 by reference in /xxx/xxx/xxx/test.php on line 36. 

线路是....

$sql11->bind_param('ss',"",$Username);

怎么了？