我正在使用xampp,并且我的文件位于xampp/htdocs/Project中。 我认为PHP代码出现在网站上,因为我在浏览器中调用它的方式:http://localhost:8081/login.php但我不确定这是否是问题所在。登录页面显示的是php代码和下一步
我的login.php是像现在这样的权利:http://imgur.com/a/nCpI1
的login.php代码:
<!DOCTYPE html>
<html lang="pt-PT">
<head>
<meta charset="UTF-8">
<title>LOGIN</title>
<link rel="stylesheet" type="text/css" href="styles/menu.css">
<link rel="stylesheet" type="text/css" href="styles/homesheet.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
<link href="https://fonts.googleapis.com/css?family=Fira+Sans+Condensed" rel="stylesheet">
</head>
<body>
<?php
error_reporting(E_ALL);
session_start();
//Start Database
$con = mysqli_connect("localhost","root","","loja");
// Check connection
if (!$con) {
echo "<div>";
echo "Falha ao conectar ao MySQL: " . mysqli_connect_error();
echo "</div>";
}
if(isset($_SESSION["name"]) && $_SESSION["name"])
{
echo "You are already logged in, ".$_SESSION['name']."! <br> I'm Loggin you out M.R ..";
unset($_SESSION);
session_destroy();
header('Location: login.php');
exit;
}
$loggedIn = false;
$userName = isset($_POST["name"]) ? $_POST["name"] : null;
$userPass = isset($_POST["pass"]) ? $_POST["pass"] : null;
if ($userName && $userPass)
{
$query = "SELECT username FROM login WHERE username = '$userName' AND password = '$userPass'";
$result = mysqli_query($con, $query);
$row = mysqli_fetch_array($result);
if(!$row){
echo "<div>"
echo "Dados invalidos.";
echo "</div>"
}
else {
$loggedIn = true;
}
}
if (!$loggedIn)
{
echo "
<div style='width:500px;'>
<h3>Login</h3>
<form method='post'>
<label>Username</label>
<input type='text' name='name' placeholder='Username' value='$userName'/>
<label>Password</label>
<input type='password' name='pass' placeholder='Password' value='$userPass'/>
<button>Login</button>
</form>
</div>
<footer>
<h4 style="text-align:right;">Copyright © est.setubal.com</h4>
</footer>
";
}
else{
echo "<div>";
echo "Iniciou a sessao como: $userName!";
echo "</div>";
header('Location: home.html');
$_SESSION["name"] = $userName;
}
?>
</body>
</html>
session.php文件:
<?php
// Establishing Connection with Server by passing server_name, user_id and password as a parameter
$connection = mysql_connect("localhost", "root", "");
// Selecting Database
$db = mysql_select_db("loja", $connection);
session_start();// Starting Session
// Storing Session
$user_check=$_SESSION['login_user'];
// SQL Query To Fetch Complete Information Of User
$ses_sql=mysql_query("select user from login where user='$user_check'",$connection);
$row = mysql_fetch_assoc($ses_sql);
$login_session =$row['user'];
if(!isset($login_session)){
mysql_close($connection); // Closing Connection
header('Location: main.php'); // Redirecting To Home Page
}
?>
我的表 “登录” 作为用户名和密码字段在里面。我需要一个全新的眼睛,可以告诉我还需要做些什么才能使登录正常工作。
您是否设置了端口8081?不要使用'mysql_ *'进行新的开发。 – chris85
如果你刚刚开始使用PHP,从一些框架开始,将会得到该代码组织的肯定..例如:[CakePHP3](https://cakephp.org/)..有一个很好的博客教程做从头开始。 – fabricio
我确实使用那个端口。而且我正在使用'mysql_ *',因为我正在使用旧的项目登录名。但我尝试在PDO中搜索一个登录系统,并尝试使用它们,但它们不起作用。 – fabimetabi