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訪問會話變量時,我得到錯誤:錯誤在PHP
Notice: Undefined variable: _SESSION in C:\xampp\htdocs\vts\include\functions.php on line 53
在訪問會話變量。以下是我的代碼:
<?php
include_once('include/config.php');
class User
{
//Database connect
public function __construct(){
$db = new DB_Class();
}
//Registration process
public function register_user($fname, $lname, $username, $email, $password){
$password = md5($password);
$sql = mysql_query("SELECT u_id from vt_user WHERE u_username = '$username' or u_email = '$email'");
$no_rows = mysql_num_rows($sql);
if ($no_rows == 0){
$result = mysql_query("INSERT INTO vt_user(u_fname, u_lname, u_username, u_email, u_password) values ('$fname', '$lname', '$username', '$email','$password')") or die(mysql_error());
return $result;
}else{
return FALSE;
}
}
// Login process
public function check_login($emailusername, $password)
{
$password = md5($password);
$result = mysql_query("SELECT u_id from vt_user WHERE u_email = '$emailusername' and u_password = '$password'");
$user_data = mysql_fetch_array($result);
$no_rows = mysql_num_rows($result);
if ($no_rows == 1){
$_SESSION['login']=TRUE;
$_SESSION['uid'] = $user_data['uid'];
return TRUE;
}else{
return FALSE;
}
}
// Getting name
public function get_session_details()
{
if(isset($_SESSION['login'])){
$id = $_SESSION['uid'];
$result = mysql_query("SELECT u_id, u_fname, u_lname, u_username, u_email from vt_user where u_id = $id");
$user_data = mysql_fetch_array($result);
echo "Hello ".$user_data['name'];
}
}
// Getting session
public function get_session()
{
return $_SESSION['uid'];
}
// Logout
public function user_logout()
{
$_SESSION['login'] = FALSE;
session_destroy();
}
}
?>
有沒有其他方法可以做到這一點?我如何在get_session()
函數中訪問這個變量?
您是否添加了'session_start()'? – Bram
注意:'mysql'擴展名自PHP 5.5.0起棄用,並且將來會被刪除。相反,應該使用MySQLi或PDO_MySQL擴展。 – GolezTrol