Question

我使用$_SESSION['name'];后使用Notice: Undefined index: name in C:\Users\..\logged_in.php on line 2从数据库中呼叫姓名我得到这个Hey, <?php echo $_SESSION['name']; ?>. You are logged in. Try to close this browser tab and open it again. Still logged in! ;) <a href="index.php?logout">Logout</a>你能告诉我什么了吗？请帮忙...

登录in.php

<?php

/**
 * Class login
 * handles the user's login and logout process
 */
class Login
{
    /**
     * @var object The database connection
     */
    private $db_connection = null;
    /**
     * @var array Collection of error messages
     */
    public $errors = array();
    /**
     * @var array Collection of success / neutral messages
     */
    public $messages = array();

    /**
     * the function "__construct()" automatically starts whenever an object of this class is created,
     * you know, when you do "$login = new Login();"
     */
    public function __construct()
    {
        // create/read session, absolutely necessary
        session_start();

        // check the possible login actions:
        // if user tried to log out (happen when user clicks logout button)
        if (isset($_GET["logout"])) {
            $this->doLogout();
        }
        // login via post data (if user just submitted a login form)
        elseif (isset($_POST["login"])) {
            $this->dologinWithPostData();
        }
    }

    /**
     * log in with post data
     */
    private function dologinWithPostData()
    {
        // check login form contents
        if (empty($_POST['user_name'])) {
            $this->errors[] = "Username field was empty.";
        } elseif (empty($_POST['user_password'])) {
            $this->errors[] = "Password field was empty.";
        } elseif (!empty($_POST['user_name']) && !empty($_POST['user_password'])) {

            // create a database connection, using the constants from config/db.php (which we loaded in index.php)
            $this->db_connection = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

            // change character set to utf8 and check it
            if (!$this->db_connection->set_charset("utf8")) {
                $this->errors[] = $this->db_connection->error;
            }

            // if no connection errors (= working database connection)
            if (!$this->db_connection->connect_errno) {

                // escape the POST stuff
                $user_name = $this->db_connection->real_escape_string($_POST['user_name']);

                // database query, getting all the info of the selected user (allows login via email address in the
                // username field)
                $sql = "SELECT user_name, user_email, user_password_hash
                        FROM users
                        WHERE user_name = '" . $user_name . "' OR user_email = '" . $user_name . "';";
                $result_of_login_check = $this->db_connection->query($sql);

                // if this user exists
                if ($result_of_login_check->num_rows == 1) {

                    // get result row (as an object)
                    $result_row = $result_of_login_check->fetch_object();

                    // using PHP 5.5's password_verify() function to check if the provided password fits
                    // the hash of that user's password
                    if (password_verify($_POST['user_password'], $result_row->user_password_hash)) {

                        // write user data into PHP SESSION (a file on your server)
                        $_SESSION['user_name'] = $result_row->user_name;
                        $_SESSION['user_email'] = $result_row->user_email;
                        $_SESSION['user_login_status'] = 1;

                    } else {
                        $this->errors[] = "Wrong password. Try again.";
                    }
                } else {
                    $this->errors[] = "This user does not exist.";
                }
            } else {
                $this->errors[] = "Database connection problem.";
            }
        }
    }

    /**
     * perform the logout
     */
    public function doLogout()
    {
        // delete the session of the user
        $_SESSION = array();
        session_destroy();
        // return a little feeedback message
        $this->messages[] = "You have been logged out.";

    }

    /**
     * simply return the current state of the user's login
     * @return boolean user's login status
     */
    public function isUserLoggedIn()
    {
        if (isset($_SESSION['user_login_status']) AND $_SESSION['user_login_status'] == 1) {
            return true;
        }
        // default return
        return false;
    }
}

的login.php

{{1}}

Answer 1

也许无法创建会话。您在文件中使用了start_session()吗？

echo用户名，如<?php echo (isset($_SESSION['name'])) ? $_SESSION['name'] : 'guest'; ?>，并查看是否已创建会话。

Answer 2

我在session_start函数中看到了Login::__construct，但在尝试访问new Login个变量之前，您是否构建了$_SESSION？我没有看到......我建议确保session_start()在您的初始化代码中的某处，在其他任何地方之前调用。

Answer 3

我假设有一些其他代码调用登录脚本。无论如何，该通知意味着该名称＆＃39; key没有在$ _SESSION中注册，这是有道理的，因为我认为你的意思是＆＃39; user_name＆＃39;。

所以试试：

Hey, <?php echo $_SESSION['user_name']; ?>. You are logged in.
Try to close this browser tab and open it again. Still logged in! ;)

<a href="index.php?logout">Logout</a>

否则你必须注册＆＃39;名称＆＃39;在会话中，假设$ result_row具有name属性：

if (password_verify($_POST['user_password'], $result_row->user_password_hash)) {
    // write user data into PHP SESSION (a file on your server)
    $_SESSION['name'] = $result_row->name; 
    $_SESSION['user_name'] = $result_row->user_name;
    $_SESSION['user_email'] = $result_row->user_email;
    $_SESSION['user_login_status'] = 1;

}

Answer 4

1st。请将您的session_start()移到一个文件中，该文件将包含在您的所有php文件中，删除来自您的Login类：）

也许您可以将其移动到将常量设置为DB_HOST，DB_USER等的位置。

第二次检查您的$ _SESSION密钥:)您正在设置$_SESSION['user_name']，但尝试抓住$_SESSION['name']

Answer 5

您无法在不启动会话的情况下访问$_SESSION变量。此外，如果您已经在php文件中的任何地方启动过会话，那么您将收到一条会话已经启动的警告。因此，在访问$_SESSION变量之前，请考虑使用以下语句：

if( !session_id() ) session_start();

注意：未定义的索引：第2行的C：\ Users \ .. \ logged_in.php中的名称

5 个答案: