我通过登录表单一直收到错误的用户名和密码

Question

我无数次扫描了我的代码,根本无法解决这个问题.每次我点击登录,它都会返回错误的密码和用户名.

admin.php的

<div class="container">
    <div class="row">
        <div class='col-md-3'></div>
        <div class="col-md-6">
            <div class="login-box well">
                    <form action="connectivity.php" method="post">
                        <legend>Sign In</legend>
                        <div class="form-group">
                            <label for="user">Username</label>
                            <input type="text" id="user" name="user" placeholder="Username"class="form-control" />
                        </div>
                        <div class="form-group">
                            <label for="pass">Password</label>
                            <input type="text" id="pass" name="pass" placeholder="Password" class="form-control" />
                        </div>
                        <div class="form-group">
                            <a href="index.php" class="btn btn-link pull-right">Return to Login</a>
                            <input name="submit" type="submit" class="btn btn-warning btn-login-submit btn-block m-t-md" value="Login" />
                        </div>
                    </form>

            </div>
        </div>
        <div class='col-md-3'></div>
    </div>
</div>

connectivity.php

<?php 
session_start();

define('DB_HOST', 'localhost'); 
define('DB_NAME', 'login-invoices'); 
define('DB_USER','root'); 
define('DB_PASSWORD',''); 

$con=mysql_connect('localhost', 'username','password', '') or die("Failed to connect to MySQL: " . mysql_error()); $db=mysql_select_db(DB_NAME,$con) or die("Failed to connect to MySQL: " . mysql_error()); 
/*
$ID = $_POST['user']; 
$Password = $_POST['pass']; 
*/
function SignIn() 
{ 
    session_start(); //starting the session for user profile page 
    if(!empty($_POST['user'])) //checking the 'user' name which is from Sign-In.html, is it empty or have some text 
    { 
        $user = $_POST[user];
        $pass = $_POST[pass];

            $query = "SELECT * FROM login where username = '$user' AND password = '$pass'";
            $query = mysql_query($query) or die(mysql_error()); 
        $row = mysql_fetch_array($query) or die(mysql_error());
        if(!empty($row['user']) AND !empty($row['pass'])) 
        {  
            header("location: Home.html");

        } else { 
            echo "SORRY... YOU ENTERD WRONG ID AND PASSWORD... PLEASE RETRY..."; 
            } 
        } 
    } 
    if(isset($_POST['submit'])) 
    { 
        SignIn(); 
    } 

?>

这是我的数据库的图片

这是表格设置

Answer 1

你有一些缺失的报价:

$user = $_POST[user];
$pass = $_POST[pass];

固定版本:

$user = $_POST['user'];
$pass = $_POST['pass'];

正如Chris所说,你的列名在这里是不正确的:

 if(!empty($row['user']) AND !empty($row['pass'])) 

根据你的SQL,它们应该是:

 if(!empty($row['username']) AND !empty($row['password'])) 

最后,快速在线搜索"SQL Inject Attacks":

$query = "SELECT * FROM login where username = '$user' AND password = '$pass'";

如果我使用以下用户名登录:

'; SELECT * FROM login LIMIT 1; --

我很确定我可以访问你的系统!