加密密码登录问题


Encrypt password login issue

我在注册过程中使用以下代码加密用户注册密码。但问题是,我无法再次使用相同的密码登录,我可能是因为数据库中的密码不同且加密,与用户输入的密码不同。

<?php 
if(isset($_POST['submit'])) {
    $firstname = $_POST['firstname'];
    $lastname = $_POST['lastname'];
    $username = $_POST['username'];
    $email = $_POST['email'];
    $password = $_POST['password'];
    if(!empty($firstname) && !empty($lastname) && !empty($username) && !empty($email) && !empty($password)) {
         $firstname = mysqli_real_escape_string($db_connect, $firstname);
         $lastname  = mysqli_real_escape_string($db_connect, $lastname);
         $username  = mysqli_real_escape_string($db_connect, $username);
         $email     = mysqli_real_escape_string($db_connect, $email);
         $password  = mysqli_real_escape_string($db_connect, $password);
         $sql = "SELECT randsalt FROM user ";
         $select_randsalt_query = mysqli_query($db_connect, $sql);
         if(!$select_randsalt_query) {
             die("Query failed".mysqli_error($db_connect));
         }
         while($row = mysqli_fetch_array($select_randsalt_query)) {
             $salt = $row['randsalt'];
             ///crypt function takes 2 parameter. one from DB 
             ///and other from user input. 
             // $password = crypt($password, $salt);
         }
         $sql_register ="INSERT INTO user(user_firstname, user_lastname, username, user_email, user_password, user_role )";
         $sql_register .="VALUES('{$firstname}', '{$lastname}', '{$username}', '{$email}', '{$password}', 'Unknown' ) ";
         $query_register = mysqli_query($db_connect, $sql_register);
         if(!$query_register) {
             die("Query failed".mysqli_error($db_connect));
         }
         $message = "<h3>Your Registration has been Submitted</h3>";
     } else {
         $message = "<h3>You Can't leave field Empty</h3>";
     }
 } else {
    $message = '';
 }
 ?>

我试着在login.php 中做这样的事情

<?php 
if(isset($_POST['submit'])){
    $Username = $_POST['Username'];
    $Password = $_POST['Password'];
    //To prevent SQL injection and store into new variable
    $Username = mysqli_real_escape_string($db_connect, $Username);
    $Password = mysqli_real_escape_string($db_connect, $Password);
    $sql_login = "SELECT * FROM user WHERE username = '{$Username}' ";
    $query_login = mysqli_query($db_connect, $sql_login);
        if(!$query_login){
            die("Query Failed".mysqli_error($db_connect));
        }

    while($row = mysqli_fetch_assoc($query_login)){
            $username       = $row['username'];
            $user_password  = $row['user_password'];
            $user_firstname = $row['user_firstname'];
            $user_lastname  = $row['user_lastname'];
            $user_email     = $row['user_email'];
            $user_role      = $row['user_role'];
    }
            $Password = crypt($Password, $user_password);
    ///User validation
    if( ($Username === $username && $Password === $user_password) && $user_role === "Admin"){
            //Using session to store information from db
            //Using session from right to left. Right is the variable got from db.
            $_SESSION['USERNAME'] = $username;
            $_SESSION['PASSWORD'] = $user_password ;
            $_SESSION['FIRSTNAME'] = $user_firstname;
            $_SESSION['LASTNAME'] = $user_lastname;
            $_SESSION['EMAIL'] = $user_email;
            $_SESSION['ROLE'] = $user_role;
            header("Location: ../admin/index.php");

        }else{  
            header("Location: ../index.php");
        }
}
?>

但这不起作用。很抱歉,我刚进入PHP世界,对它没有深入的了解。

欢迎使用PHP开发。让我让你的生活更轻松:

  1. 不管你的教程/书/朋友说了什么,都不要转义字符串,而是使用准备好的语句。它们更容易安全地实施,你的生活也变得轻松多了。(如果你依赖于转义,并且你记得转义用户可以控制的3000个参数中的2999个,那么你仍然很脆弱。)
  2. 与其使用crypt(),不如使用password_hash()password_verify()

到处都有更新的指南,可以解释如何更好地使用这些功能,但是http://www.phptherightway.com是社区最关心的问题。