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我試圖讓我的查詢中的行數,但它不斷給我這個錯誤檢索行NUM中的mysqli
試圖讓行非對象的特性15
<?php require_once("./includes/Utilities.php") ;?>
<?php require_once("./includes/db_connection.php");?>
<?php
if(isset($_POST['submit'])){
$username = trim($_POST['username']);
$password = trim($_POST['password']);
$conf_pass = trim($_POST['conf_password']);
$email = trim($_POST['email']);
require_once('./includes/CheckPassword.php');
$check_pwd = new Db_CheckPassword($password);
$check_pwd =$check_pwd->check();
require_once('./includes/CheckUsername.php');
$sql = "SELECT * FROM accounts WHERE username = {$username}";
$result = $conn->query($sql);
$numRows = $result->num_rows;
}
?>
這是專門線15
$numRows = $result->num_rows;
爲什麼會返回一個錯誤?
當您使用MySQLi時,您並未使用準備好的語句!這導致舊版數據庫驅動程序引入了相同類型的SQL注入漏洞! –
mysqli存儲過程與執行查詢有什麼區別? – KyelJmD
我沒有說存儲過程,我說** [準備語句](http://il.php.net/manual/en/mysqli.quickstart.prepared-statements.php)** –