php - 存储 session 变量时出现问题

Question

我有一个登录脚本，当前存储 2 个变量:一个有效变量和一个用户名变量。我现在尝试添加名称变量，因此我更改了 MySQL 查询以从数据库获取名称，并尝试将名称存储在 session 变量中，但由于某种原因它只是不存储它。也许最好只是向您展示该脚本，我只学习了 2 个月的 PHP，所以我非常感谢您的帮助。

<?php
ob_start(); // Start output buffering

session_start(); //must call session_start before using any $_SESSION variables3
$_SESSION['username'] = $username;

function validateUser($username)
{

    session_regenerate_id (); //this is a security measure
    $_SESSION['valid'] = 1;
    $_SESSION['username'] = $username;
    $_SESSION['name'] = $userData['name'];
}

      $username = isset($_POST['username'])?$_POST['username']:'';
     $password = isset($_POST['password'])?$_POST['password']:'';

//connect to the database here

$hostname_Takeaway = "localhost";
$database_Takeaway = "diningtime";
$username_Takeaway = "root";
$password_Takeaway = "root";
$Takeaway = mysql_pconnect($hostname_Takeaway, $username_Takeaway, $password_Takeaway) or trigger_error(mysql_error(),E_USER_ERROR); 
mysql_select_db($database_Takeaway, $Takeaway);

$username = mysql_real_escape_string($username);

$query = "SELECT name, password, salt FROM admin_users WHERE username = '$username';";

$result = mysql_query($query) or die(mysql_error());

if(mysql_num_rows($result) < 1) //no such user exists
{
    header('Location: http://localhost/diningtime/admin-home.php?login=fail');

    die();
}
$userData = mysql_fetch_array($result, MYSQL_ASSOC);
$hash = hash('sha256', $userData['salt'] . hash('sha256', $password) );
if($hash != $userData['password']) //incorrect password


{
    header('Location: http://localhost/diningtime/admin-home.php?login=fail');

    die();
}
else
{

   validateUser($username); //sets the session data for this user
}
//redirect to another page or display "login success" message
header('Location: http://localhost/diningtime/main');
die();




//redirect to another page or display "login success" message


?>

Answer 1

您的 validateUser() 函数作用域内没有 $userData 变量，因此您将 NULL 分配给 $_SESSION['name']。

要么使 $userData 成为全局变量，以便它在函数范围内可见，要么将其作为参数传递:

function validateUser($user, $userData) {
                             ^^^^^^^^^-- pass as arg
   global $userData;
   ^^^^^^^^^^^^^^^^^--- bring var in-scope
   ...
   $_SESSION['name'] = $GLOBALS['userData']['name'];
                       ^^^^^^^^^^^^^^^^^^^^^^^^^^^^--- refer to global scope
}

这 3 个选项中的任何一个都可以解决问题(只是不要同时执行所有三个选项)