PHP登录: does not respond well

Question

这是一个简单的用户登录。我只是想了解这个概念。让我解释一下我的代码问题。这是我的表单:

<form action = "check_login.php" action = "POST">
    User:<input type="text" name="user"><br/>
    Password:<input type="password" name="password"><br/>
    <input type="submit" name="submit" value="Sign In">
</form>

check_login.php:

    <?php
    error_reporting(E_ALL);
    $username = $_POST['user'];
    $password = $_POST['password'];

    if($username && $password) {
        $host = 'localhost';    
        $user = 'root';

        $con = mysql_connect($host, $user, '') or die("Couldn't Connect");
        mysql_select_db('first_site');

        $sql = "SELECT * FROM user WHERE user_name= '$username' and user_password ='$password'";
        $query = mysql_query($sql, $con);

        $count = mysql_num_rows($query);
        echo $count;

     }
     else {
        die "Enter username and password";
     }

我维护了一个数据库first_site和表user。该表有 3 列:id、user_name 和 user_password。

如果我输入用户名和密码，$count 变量应该保存值 1，但它恰好保存 0。我一无所知。帮助!!

Answer 1

代码本身是危险的，你没有使其安全以防止注入(inject)'

 <?php
error_reporting(E_ALL);
// Escaped For SQL Injection Prevention
$username = mysql_real_escape_string($_POST['user']);
$password = mysql_real_escape_string($_POST['password']);

if($username && $password) {
    $host = 'localhost';    
    $user = 'root';

    $con = mysql_connect($host, $user, '') or die("Couldn't Connect");
    mysql_select_db('first_site');

    $sql = "SELECT user_name FROM user WHERE user_name= '$username' and user_password ='$password'";
    $query = mysql_query($sql, $con);

    // If Exists
    if( $query ) {
        $count = mysql_num_rows($query);
         echo $count;
    }
    else {

        die(mysql_error());

    }

 }
 else {
    die ("Enter username and password");
 }

同时将形式更改为

<form action = "check_login.php" method = "POST">
User:<input type="text" name="user"><br/>
Password:<input type="password" name="password"><br/>
<input type="submit" name="submit" value="Sign In">

尝试上面的方法，如果有的话，它应该会提示我们一个 mysql 错误，它也比原来的更安全。