我目前正在使用PHP编写一个CMS,以便回到PHP(我总是使用它)。然而,由于某些奇怪的原因,当“包含”或“需要”我的类文件时,它只是停止php脚本,我的登录表单(login.php的html)不显示(无论我是否登录)。任何帮助?这里有两个我的脚本:PHP脚本不会运行
的login.php:
<?php
session_start();
include "classes.php";
if(isset($_GET['logout'])) {
setupSession(2);
}
if($_SESSION['status'] == "online") header("location: admin.php");
if($_POST && isset($_POST['username']) && isset($_POST['password'])) {
$un = $_POST['username'];
$pwd = $_POST['password'];
$mysql = new mySql();
$mysql->validateUser($un, $pwd);
} else $attempt = 2;
?>
<html>
<head>
<title>Log In</title>
</head>
<body>
<form method="post" action="">
<label for="username">username: </label>
<input type="text" name="username" />
<label for="password">password: </label>
<input type="password" name="password" />
<input type="submit" value="Log In" name="submit" />
</form>
</body>
</html>
和classes.php
<?php
class mySql {
protected $dbname;
protected $dbuser;
protected $dbpass;
protected $db;
private $conn;
function __construct() {
$conn = new mysqli($dbname, $dbuser, $dbpass, $db);
}
public function validateUser($username, $password) {
$query = "SELECT * FROM users WHERE username = ? AND password = ? LIMIT 1";
if($stmt = $this->conn->prepare($query)) {
$stmt->bind_param('ss', $username, $password);
$stmt->execute();
if($stmt->fetch()) {
$stmt->close();
setupSession(1);
} else $attempt = 1;
}
}
}
function setupSession($status) {
switch($status) {
case 1:
$_SESSION['status'] = "online";
//other user variables
header("location: admin.php");
break;
case 2:
unset($_SESSION['status']);
if(isset($_COOKIE[session_name()])) {
setcookie(session_name(), '', time() - 1000);
}
session_destroy();
break;
default:
session_start();
if($_SESSION['status'] != "online") header("location: login.php");
break;
}
}
?>
你的'error_reporting'开启了吗? – 2010-06-29 17:04:57
另外请注意,你必须,必须在'header(“location:...”)之后**'die()',否则受保护的内容将被发送到文档正文中的客户端。 – 2010-06-29 17:05:25
不确定关于error_reporting,我正在使用MAMP进行开发。对于死亡(),谢谢。 – user379229 2010-06-29 22:45:50