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
$stmt中的error属性为空,我没有真正的问题,但我只是想知道是什么原因导致出现此警告消息.
> UPDATE query with prepared statements
> http://php.net/manual/en/mysqli-stmt.param-count.php
您的MysqL连接可能尚未建立.在MysqLi :: __ construct()之后你必须检查
原文链接：https://www.f2er.com/php/137071.htmlmysqli::$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()); }
