注意:未定义的索引:第2行的C:\ Users \ .. \ logged_in.php中的名称
i'm calling name from database using $_SESSION['name'];
after using this i'm getting this Notice: Undefined index: name in C:\Users\..\logged_in.php on line 2
can you tell me what's goin on? please help please...
logged-in.php
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>
login.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;
}
}
I'm assuming that there is some other code calling the login script. Anyways, that notice means that the 'name' key is not registered in the $_SESSION, which makes sense since I think you mean 'user_name'.
so try:
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>
Otherwise you have to register the 'name' in the session, assuming $result_row has a name attribute:
if (password_verify($_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;
}
I see session_start
in your Login::__construct
function, but do you construct new Login
before attempting to access $_SESSION
variables? I'm not seeing that... I would suggest making sure session_start()
is somewhere in your init code, called before anything else.
Maybe the session can't be created. Did you use start_session()
on your file?
echo
username like that <?php echo (isset($_SESSION['name'])) ? $_SESSION['name'] : 'guest'; ?>
and see if session has been created.
1st. Please move your session_start()
to a file which will be included in all your php files and REMOVE it from your Login class :)
Perhaps you can move it to where you set constants as DB_HOST
, DB_USER
and etc..
2nd Check your $_SESSION keys :) you are setting $_SESSION['user_name']
but try to catch $_SESSION['name']
You can't access $_SESSION
variable without starting a session. Also if you have already started the session anywhere before in your php file, then you will get warning that a session is already started. so consider using the following statement, before you access $_SESSION
variable:
if( !session_id() ) session_start();