Question

我目前正在使用PHP编写CMS，以便重新使用PHP（我总是使用它）。但是，由于一些奇怪的原因，当“包含”或“要求”我的类文件时，它只是停止php脚本，我的登录表单（login.php的html）没有显示（我是否登录）。有帮助吗？这是我的两个脚本：

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;
    }
}

?>

Answer 1

您有范围问题。

$conn = mysqli(....)

应为$this->conn = mysqli(....)

Answer 2

所需脚本破坏父级的原因不多：所需文件不存在，出错或调用 exit（）或 die（）< / em>的。

您确定文件classes.php与脚本位于同一文件夹中，还是位于包含路径中？

这是您正在使用的确切代码吗？

使用这样的构造函数：

function __construct() { $conn = new mysqli($dbname, $dbuser, $dbpass, $db); }

你怎么连接数据库？

$mysql = new mySql();

Answer 3

function __construct() {
    $conn = new mysqli($dbname, $dbuser, $dbpass, $db);
}

应该

function __construct($dbname, $dbuser, $dbpass, $db) {
    $this->dbname = $dbname;
    $this->dbuser = $dbuser;
    $this->dbpass = $dbpass;
    $this->db     = $db;
    $this->connect();
}

function connect()
{
    $this->conn = new mysqli($this->dbname, $this->dbuser, $this->dbpass, $this->db);
}

那种性质。

Answer 4

error_reporting (1);

PHP脚本将无法运行

4 个答案: