在 php 登錄會話中獲取存儲的數據庫變量

Question

我想在我的個人資料頁面上讀取 $Session[] 變量，但我在嘗試時遇到了問題。 這是我的登錄文件中的 php 腳本

<?php
    if (isset($_POST['LoginBtn']))
    {
      $username = $_POST['username'];
      $password = $_POST['password'];
      if (empty($username) || empty($password))
      {
        $show -> showError("please fill out all the fields");
      }
      else
      {
        $SQLCheckUser = $odb -> prepare("SELECT COUNT(*) FROM `Account` WHERE `username` = :user AND `password` = :password LIMIT 1");
        $SQLCheckUser -> execute(array(':user' => $username, ':password' => hash('SHA512', $password)));
        $loginCheck = $SQLCheckUser -> fetchColumn(0);
        if ($loginCheck)
        {

          $SQLGetID = $odb -> prepare("SELECT `ID`, FROM `Account` WHERE `username` = :username LIMIT 1");
          $SQLGetID -> execute(array(':username' => $username));

          $_SESSION['username'] = $username;

          $_SESSION['ID'] = $SQLGetID -> fetchColumn(0);
          $show -> showSuccess('Willkommen... <meta http-equiv="refresh" content="2;url=index.php">');
        }
        else
        {
          $show -> showError('user was not found!');
        }
      }
    }
    ?>

在帳戶數據庫中，我有變量，如用戶名、密碼、名字、姓氏、地址，我想在 profile.php 頁面上顯示它們，例如像這樣

          <h4 class="profile-user"><?php echo $_SESSION['firstname']; ?></h4>
          <p class="profile-job"><?php echo $_SESSION['lastname']; ?></p>
          <p class="profile-job"><?php echo $_SESSION['adress'];?></p>

我怎么能解決這個問題？ 頁面包含一個開始會話的 php 文件

<?php session_start(); ?>

和一個數據庫連接

<?php
$show = new show();
$user = new user($odb);
$status = new status($odb);
class show
{
function showError($error)
{
    echo '<div class="alert alert-danger"><a class="close" data-                                dismiss="alert" href="#">&times;</a><h4 class="alert-heading">error!</h4>'.$error.'</div>';
}
function showSuccess($success)
{
    echo '<div class="alert alert-success"><a class="close" data-dismiss="alert" href="#">&times;</a><h4 class="alert-heading">Login success!</h4>'.$success.'</div>';
}
}

class user
{
var $odb;

function __CONSTRUCT($odb)
{
    $this -> odb = $odb;
}
function loggedIn()
{
    if (isset($_SESSION['username'], $_SESSION['ID']))
    {
        return true;
    }
    else
    {
        return false;
    }
}
function isAdmin()
{
    $SQL = $this -> odb -> prepare("SELECT `admin` FROM `Account` WHERE `ID` = :id");
    $SQL -> execute(array(':id' => $_SESSION['ID']));
    $rank = $SQL -> fetchColumn(0);
    if ($rank == 1)
    {
        return true;
    }
    else
    {
        return false;
    }
    }
}`

我的目標是在個人資料頁面上顯示登錄用戶的所有數據庫變量（全名、地址、電話號碼），如果有人可以幫助我修復我的代碼，將不勝感激:)

Answer 1

這個頁面 profile.php

session_start();
$dbh = new PDO(" --- connection string --- ");
$user = new user($dbh);
if ($user->loggedIn()) {
    $stmt = $odb->prepare("SELECT `firstname`, `lastname`,`adress`,FROM `Account` WHERE ID=$_SESSION['ID'] LIMIT 1");
    $stmt->execute();
    $row = $stmt->fetch();
}
foreach ($row as $item)
{
    <h4 class="profile-user"><?php echo $item['firstname']; ?></h4>
    <p class="profile-job"><?php echo $item['lastname']; ?></p>
    <p class="profile-job"><?php echo $item['adress']; ?></p>
} 

Answer 2

我收到此解決方案的錯誤，我通過將 profile.php 中的代碼更改為此進行了修復。

<div class="col-lg-3">
      <!-- Page Widget -->

     <?php 
     $stmt = $odb->prepare("SELECT 'vorname', 'nachname','email',FROM 'Account' WHERE ID=$_SESSION[ID] LIMIT 1");
     $stmt->execute();
     $row = $stmt->fetch();

     foreach ((array) $row as $item){
       $vorname = $item['vorname']; // german word for firstname
       $nachname = $item['nachname']; // german word for lastname
       $email = $item['email']; // german word for email

     } ?>
      <h4 class="profile-user"><?php echo $vorname ?></h4>

但如果我重新加載屏幕，它似乎不會出現。

對我來說看起來很奇怪