我在MAMP工作,尝试登录功能 我的连接代码是:
$servername = "localhost";
$username = "root";
$password = "root";
$db = "world";
$mysqli = new mysqli($servername, $username, $password, $db);
if($mysqli->connect_error){
die("Connection failed: " . $conn->connect_error);
}
我的登录功能:
if (isset($_POST['email'], $_POST['p'])){
$email = $_POST['email'];
$password = $_POST['p']; //hashed password
if(login($email, $password, $mysqli) == true){
header('Location: ../protected_page.php');
}else{
echo 'failed login';
}
}
function login($email, $password, $mysqli){
if($stmt = $mysqli->prepare("SELECT USERID, USERNAME, PASSWORD, SALT FROM USERS WHERE EMAIL = ? LIMIT 1")){
$stmt = $mysqli->bind_param('s', $email);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($user_id, $username, $db_password, $salt);
$stmt->fetch();
}
错误:
[2015年10月17日08:46:06欧洲/柏林] PHP致命错误:在第24行的/Users.../Site/include/functions.php中调用未定义的方法mysqli :: bind_param()< / p>
答案 0 :(得分:1)
http://php.net/manual/en/mysqli-stmt.bind-param.php
从手册中可以看出,这不是mysqli对象的方法,而是mysqli_stmt对象的方法。
运行$stmt = $mysqli->bind_param('s', $email);
答案 1 :(得分:0)
你需要为你的陈述打电话bind_param
,如下所示:
$stmt->bind_param('s', $email);
您之前已经定义了$stmt
一行。 $mysqli
没有bind_param
功能。