Question

我正在尝试获取当前登录用户的用户名，以便我可以执行诸如在欢迎屏幕上显示用户名或检查该用户是否有权访问页面等内容。

$ logedInUsername返回“Array”，我原本认为它返回“Bob”，因为那是我登录的用户。

的index.php

<?php
// Start Require Login
require("common.php");
if(empty($_SESSION['user']))
{
    header("Location: login.php");
    die("Redirecting to login.php");
}

// get current logged in user   
$logedInUsername = $_SESSION['user'];
echo $logedInUsername;

// check if the username is equal to admin
if($logedInUsername == "admin")
{
  echo "You are a admin!";
} 
else 
{
  echo "You are NOT a admin!";
}
// End Require Login

// ... html code below here ...

的common.php

<?php 
$username = "username"; 
$password = "password"; 
$host = "localhost"; 
$dbname = "db"; 

$options = array(PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8'); 

try 
{ 
    $db = new PDO("mysql:host={$host};dbname={$dbname};charset=utf8", $username, $password, $options); 
} 
catch(PDOException $ex) 
{ 
    die("Failed to connect to the database: " . $ex->getMessage()); 
} 

$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 

$db->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC); 

if(function_exists('get_magic_quotes_gpc') && get_magic_quotes_gpc()) 
{ 
    function undo_magic_quotes_gpc(&$array) 
    { 
        foreach($array as &$value) 
        { 
            if(is_array($value)) 
            { 
                undo_magic_quotes_gpc($value); 
            } 
            else 
            { 
                $value = stripslashes($value); 
            } 
        } 
    } 

    undo_magic_quotes_gpc($_POST); 
    undo_magic_quotes_gpc($_GET); 
    undo_magic_quotes_gpc($_COOKIE); 
} 

header('Content-Type: text/html; charset=utf-8'); 

session_start();

的login.php

<?php 
require("common.php"); 

$submitted_username = ''; 

if(!empty($_POST)) 
{ 
    $query = " 
        SELECT 
            id, 
            username, 
            password, 
            salt, 
            email 
        FROM users 
        WHERE 
            username = :username 
    "; 

    $query_params = array( 
        ':username' => $_POST['username'] 
    ); 

    try 
    { 
        $stmt = $db->prepare($query); 
        $result = $stmt->execute($query_params); 
    } 
    catch(PDOException $ex) 
    {  
        die("Failed to run query: " . $ex->getMessage()); 
    } 

    $login_ok = false; 


    $row = $stmt->fetch(); 
    if($row) 
    { 

        $check_password = hash('sha256', $_POST['password'] . $row['salt']); 
        for($round = 0; $round < 65536; $round++) 
        { 
            $check_password = hash('sha256', $check_password . $row['salt']); 
        } 

        if($check_password === $row['password']) 
        { 
            $login_ok = true; 
        } 
    } 


    if($login_ok) 
    { 
        unset($row['salt']); 
        unset($row['password']); 

        $_SESSION['user'] = $row; 

        header("Location: index.php"); 
        die("Redirecting to: index.php"); 
    } 
    else 
    { 
        print("Login Failed."); 

        $submitted_username = htmlentities($_POST['username'], ENT_QUOTES, 'UTF-8'); 
    } 
} 

?> 
<h1>Login</h1> 
<form action="login.php" method="post"> 
    Username:<br /> 
    <input type="text" name="username" value="<?php echo $submitted_username; ?>" /> 
    <br /><br /> 
    Password:<br /> 
    <input type="password" name="password" value="" /> 
    <br /><br /> 
    <input type="submit" value="Login" /> 
</form>

Answer 1

试试这个。

<强>的login.php

$_SESSION['user'] = $row['username'];

Answer 2

您将整行设置为$_SESSION['user']，因此它是一个数组。

您可以替换

$logedInUsername = $_SESSION['user'];;

与

$logedInUsername = $_SESSION['user']['username'];

PHP / MySQL获取当前登录用户的用户名

2 个答案: