Question

所以我想创建一个社交网络并从codeetastic跟踪一些youtube视频，但在第四部电影中，我的登录功能没有做出正确的反应。你们有什么想法吗？（的index.php）

<?php include "templates/nav.php"; ?>
<!DOCTYPE html>
<html>
    <head>
        <meta charset="UTF-8"/>
        <title>EverTime</title>
        <link rel="stylesheet" href="css/main.css"/>
        <script src="js/main.js"></script>
    </head>
    <body>
        <div id="middle">
            <h1 id="welcome">Welcome to EverTime!</h1>
            <form action="parse/login.php" method="post">
                <input type="text" name="username" id="username" placeholder="Username..."/>
                <input type="password" name="password" id="password" placeholder="Password..."/>
                <input type="submit" name="submit" id="submit" value="Log in"/><a id="link" href="register.php">Create an account</a>
            </form>
        </div>  
    </body>
</html>

（login.php中）

<?php
//connect
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("evertime") or die(mysql_error());

//login script
    if(isset($_POST['username'])) {
        $username = $_POST['username'];
        $password = $_POST['password'];

        $md5pass = md5($password);

        $query = mysql_query("SELECT * FROM `users` WHERE username='$username' AND password='$md5pass'");
        $get = mysql_fetch_assoc($query);

        if($get = ""){
            echo "User does not exist";
        } else {
            echo "User does exist";
        }
    }
?>

（config.php中）

<?php

//connect to database
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("evertime") or die(mysql_error());
?>

顺便说一下，我正在使用xampp和php 5.4.15以及mysql（不是mysql * i *）

Answer 1

    if($get = ""){

应该是

    if($get == ""){

我的MySql，php组合无法正常工作

1 个答案: