php – 使用预准备语句时“不允许属性访问”警告

我正在尝试使用AES_ENCRYPT()对我的密码进行编码来登录系统.但是在尝试实现这些代码时,我从xdebug得到了一些警告：

...
$key = 'd0gis=SUPER-cute';
$sql = "SELECT * FROM `users2` WHERE username = ? AND pwd = AES_ENCRYPT(?,?)";
$stmt = $conn->stmt_init();
$stmt->prepare($sql);
$stmt->bind_param('sss',$username,$password,$key);
$stmt->execute();
$stmt->store_result();
...

当调试器遇到第8行或$stmt-> prepare($sql);时,来自xdebug的6个相同的警告表说：

(!) Warning: main(): Property access is not allowed yet in D:\xampp\htdocs\learnPHP\includes\authenticate_MysqLi.inc.PHP on line 8

$stmt中的error属性为空,我没有真正的问题,但我只是想知道是什么原因导致出现此警告消息.

用Google搜索此警告消息但未找到任何解决方案：

> UPDATE query with prepared statements
> http://php.net/manual/en/mysqli-stmt.param-count.php

您的MysqL连接可能尚未建立.在MysqLi :: __ construct()之后你必须检查 mysqli::$connect_error,这对于某些PHP版本是破坏的：

The MysqLi->connect_error property only works properly as of PHP versions 5.2.9 and 5.3.0. Use the MysqLi_connect_error() function if compatibility with earlier PHP versions is required.

请参阅mysqli::__construct()文档中的连接锅炉板：

$MysqLi = new MysqLi('localhost','my_user','my_password','my_db');

/*
 * This is the "official" OO way to do it,* BUT $connect_error was broken until PHP 5.2.9 and 5.3.0.
 */
if ($MysqLi->connect_error) {
    die('Connect Error (' . $MysqLi->connect_errno . ') '
            . $MysqLi->connect_error);
}

/*
 * Use this instead of $connect_error if you need to ensure
 * compatibility with PHP versions prior to 5.2.9 and 5.3.0.
 */
if (MysqLi_connect_error()) {
    die('Connect Error (' . MysqLi_connect_errno() . ') '
            . MysqLi_connect_error());
}

原文地址：https://www.jb51.cc/php/137071.html