檢查用戶是否登錄

Question

我已按照本教程進行操作，以便為內容管理站點構建登錄系統。 http://www.formget.com/login-form-in-php/ 。 我希望以某種方式設置站點，以便用戶需要登錄才能查看該站點，並且如果用戶通過URL輸入站點名稱，則他們將立即定向到登錄頁面。

當前，當我直接進入頁面時，出現錯誤提示：

注意：未定義的索引：第6行的C：\\ xampp \\ htdocs \\ WebDevelopment \\ V18 \\ CMS \\ session.php中的login_user

使用的文件：

login.php

<?php
session_start();
$error = ''; // variable to store error message
if (isset($_POST['login'])) {
if (empty($_POST['username']) || empty($_POST['password'])) {
    $error = "Username or Password is invalid";
}
else {
    // Define username and password
    $username = $_POST['username'];
    $password = $_POST['password'];

    $connection = mysql_connect("localhost", "root", "");
    $username = stripslashes($username);
    $password = stripslashes($password);
    $username = mysql_real_escape_string($username);
    $password = mysql_real_escape_string($password);

    // select DB
    $db = mysql_select_db("v18_apartments", $connection);
    $query = mysql_query("Select * from login where password = '$password' AND username = '$username'", $connection);
    $rows = mysql_num_rows($query);
    if ($rows == 1) {
        $_SESSION['login_user']=$username; // initializing session
        header("location: CMS-home.php");
    } else {
        $error = "Username or Password is invalid";
    }
    mysql_close($connection);
    }
  }
  ?>

index.php

<?php 
include ('login.php');

if (isset($_SESSION['login_user'])){
header("location: CMS-home.php");
}
?>

<!DOCTYPE html>
<html>
<head>
    <meta charset="UTF-8">
    <title> V18 - Login</title>
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <meta name="format-detection" content="telephone=no">
    <link rel="stylesheet" type="text/css" href="css/styles.css"/>
</head>

<body>
<div id="wrapper">
        <div class="center">
            <div id="login-form">
                <form action="" method="POST">
                    <input type="text" name="username" placeholder="username">
                    <input type="password" name="password" placeholder="********">
                     <button type="submit" name="login">Submit</button>
                    <h2><?php echo $error; ?></h2>
                </form>

            </div>
    </div>
</div>
</body>
</html>

session.php

<?php
$connection = mysql_connect("localhost", "root", "");
$db = mysql_select_db("v18_apartments", $connection);
session_start();

$user_check=$_SESSION['login_user'];
$ses_sql=mysql_query("select username from login where username =  '$user_check'", $connection);
$row = mysql_fetch_assoc($ses_sql);
$login_session = $row['username'];
if (!isset($login_session)) {
mysql_close($connection); 
header('Location : index.php');
}
?>

logout.php

<?php
session_start();
if (session_destroy()) // destroy all sessions
{
    header("Location: index.php");
}
?>

Answer 1

這個怎么樣？

session.php

<?php
session_start();

if(!isset($_SESSION['login_user']))
    header('Location : index.php');
    exit;
?>

檢查用戶是否登錄

問題描述

1 個解決方案

解決方案1
1 已采納 2015-09-10 15:14:54

檢查用戶是否登錄

問題描述

1 個解決方案

解決方案1 1 已采納 2015-09-10 15:14:54

解決方案1
1 已采納 2015-09-10 15:14:54