Question

我一直在扩展这个预先构建的php登录系统。第一个代码块是一个名为user.php的文件，我相信它会在用户登录时加载。我已经添加了它，例如

当我想在数据库中添加额外的列并将其回显时。

'depot' and  $_SESSION['depot'] = $row['depot'];

我的问题是，当我构建配置文件页面时，除非用户注销并重新登录，否则此数据不会更新。

解决此问题的好方法是什么。我是否需要删除我对user.php的添加，并在profile.php中创建一个新的SELECT语句和数组。或者我可以以某种方式刷新user.php中的变量吗？

user.php的

<?php
include('password.php');
class User extends Password{

private $_db;

function __construct($db){
    parent::__construct();

    $this->_db = $db;
}

private function get_user_hash($username){

    try {
        $stmt = $this->_db->prepare('SELECT password, username, firstname, lastname, phone, memberID, user_level, credit, email, depot FROM members WHERE username = :username AND active="Yes" ');
        $stmt->execute(array('username' => $username));

        return $stmt->fetch();

    } catch(PDOException $e) {
        echo '<p class="bg-danger">'.$e->getMessage().'</p>';
    }
}

public function login($username,$password){

    $row = $this->get_user_hash($username);

    if($this->password_verify($password,$row['password']) == 1){

        $_SESSION['loggedin'] = true;
        $_SESSION['username'] = $row['username'];
        $_SESSION['firstname'] = $row['firstname'];
        $_SESSION['lastname'] = $row['lastname'];
        $_SESSION['phone'] = $row['phone'];
        $_SESSION['memberID'] = $row['memberID'];
        $_SESSION['user_level'] = $row['user_level'];
        $_SESSION['credit'] = $row['credit'];
        $_SESSION['email'] = $row['email'];
        $_SESSION['depot'] = $row['depot'];           

        return true;
    }
}

public function logout(){
    session_destroy();
}

public function is_logged_in(){
    if(isset($_SESSION['loggedin']) && $_SESSION['loggedin'] == true){
        return true;
    }
}

}


?>

profile.php的例子

profile.php
<?php require('includes/config.php'); 

//if not logged in redirect to login page
if(!$user->is_logged_in()){ header('Location: login.php'); } 

//define page title
$title = 'Profile Page';

//include header template
require('layout/header-active.php'); 

?>

<div id="Holder">
<div class="container">       
   <div class="row">
    <div class="col-lg-6 col-lg-offset-3">
        <h1> Account Information </h1>
        <p>Username: <?php echo $_SESSION['username']; ?></p>
        <p>First Name: <?php echo $_SESSION['firstname']; ?></p>
        <p>Last Name: <?php echo $_SESSION['lastname']; ?></p>
        <p>Phone: <?php echo $_SESSION['phone']; ?></p>
        <p>Email: <?php echo $_SESSION['email']; ?></p>
        <p>Member ID: <?php echo $_SESSION['memberID']; ?></p>            

        <hr>

        <h3>Global Settings</h3>

        <form class="" action="" method="post">
            <div class="form-group">
                <label>Depot:</label> <input type="text" name="depot" class="form-control" value="<?php echo $_SESSION['depot']; ?>" />
            </div>
            <input type="submit" name="submit" value="Submit" class="btn btn-primary">
        </form> 

        <?php
        include('connect-db.php');

        if (isset($_POST['submit']))
        { 
            $memberID = $_SESSION['memberID'];
            $depot = $_POST['depot'];

            mysqli_query($conn, "UPDATE members SET depot='$depot' WHERE     memberID='$memberID'")
            or die(mysqli_error());

            header("Location: profile.php"); 
        }

        ?>

        <?php 
//include header template
        require('layout/footer.php'); 
        ?>

Answer 1

由于您的代码使用$ _SESSION变量来显示值，因此更新数据库中的值对当前会话没有影响，因为$ _SESSION仅在用户登录时更新。
如果用户再次登录，则可以看到更改，因为在您的登录功能中使用数据库中的更新信息覆盖了$ _SESSION变量。

要在不注销的情况下显示更改，请在更新数据库中的值后立即更新$ _SESSION变量。

mysqli_query($conn, "UPDATE members SET depot='$depot' WHERE     memberID='$memberID'");
$_SESSION['depot'] = $depot;

另外，你应该阅读How can I prevent SQL injection in PHP?
切勿在未正确转发的情况下将用户提交的数据直接传递给MySQL。

在php中编辑mysql并显示新结果

1 个答案: