Php 当我单击login按钮时,调用成员函数execute()时发生致命错误
当我单击登录按钮时,出现致命错误: 对非对象调用成员函数execute() 在第24行的Php 当我单击login按钮时,调用成员函数execute()时发生致命错误,php,mysql,json,Php,Mysql,Json,当我单击登录按钮时,出现致命错误: 对非对象调用成员函数execute() 在第24行的C:\wamp\www\json\login.php中发生此错误 我做了一些改变,但不起作用执行() login.php文件 <?php require("config.inc.php"); if (!empty($_POST)) { $query = "SELECT id,username,password FROM users WHERE username = :username"; $query_
C:\wamp\www\json\login.php中发生此错误
我做了一些改变,但不起作用<代码>执行()
login.php
文件
<?php
require("config.inc.php");
if (!empty($_POST)) {
$query = "SELECT id,username,password FROM users WHERE username = :username";
$query_params = array(
':username' => $_POST['username']
);
try {
$stmt = $db->prepare($query);
$result = $stmt->execute($query_params);
}
catch (PDOException $ex) {
$response["success"] = 0;
$response["message"] = "Database Error1. Please Try Again!";
die(json_encode($response));
}
$validated_info = false;
$row = $stmt->fetch();
if ($row) {
if ($_POST['password'] === $row['password']) {
$login_ok = true;
}
}
if ($login_ok) {
$response["success"] = 1;
$response["message"] = "Login successful!";
die(json_encode($response));
} else {
$response["success"] = 0;
$response["message"] = "Invalid Credentials!";
die(json_encode($response));
}
} else {
?>
<h1>Login</h1>
<form action="login.php" method="post">
Username:<br />
<input type="text" name="username" placeholder="username" />
<br /><br />
Password:<br />
<input type="password" name="password" placeholder="password" value="" />
<br /><br />
<input type="submit" value="Login" />
</form>
<a href="register.php">Register</a>
<?php
}
?>
<?php
$db = mysqli_connect('localhost','root','','json') or die(mysqli_error($connection));
?>
数据库
:-
CREATE TABLE IF NOT EXISTS `users` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`username` varchar(255) NOT NULL,
`password` varchar(255) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;
--
-- Dumping data for table `users`
--
INSERT INTO `users` (`id`, `username`, `password`) VALUES
(1, 'admin', 'admin');
看看这个
尝试使用
$stmt->bind_param(...);
您是否尝试使用示例链接实现您需要的内容?实际上我想登录,但execute()函数不起作用