通過PHP連接數據庫來檢查用戶登錄信息

Question

我寫了這段代碼，但它不工作。它基本上沒有做任何事情。這個表單放在表單所在的login.php上。表單動作設置爲php自我，因此它將信息發送給自己，我試圖檢查信息是否有效或不使用此代碼;

if ($_POST['username'] != "" && $_POST['password'] != "") { 
$data = mysqli_query($dbcon, " SELECT * 
           FROM `users` 
           WHERE `user_name` LIKE '".$_POST['username']."' 
           AND `user_password` LIKE '".$_POST['password']."'") or die(mysql_error()); 

     // if this user exists 
     if ($data->num_rows == 1) { 
       $result_row = $data->fetch_object(); 
       $_SESSION['user_name'] = $result_row->user_name; 
       $_SESSION['user_password'] = $result_row->user_password; 
       $_SESSION['user_logged_in'] = 1; 
     } else { 
      "Wrong Password"; 
     } 

} 

Answer 1

與其他人的幫助問題解決了！

謝謝大家！

if ($_POST['username'] != "" && $_POST['password'] != "") { 
$data = mysqli_query($dbcon, " SELECT * 
           FROM `users` 
           WHERE `user_name` = '".$_POST['username']."' 
           AND `user_password` = '".$_POST['password']."'") or die(mysqli_error()); 

     // if this user exists 
     if ($data->num_rows == 1) { 
       session_start(); 
       $result_row = $data->fetch_object(); 
       $_SESSION['user_name'] = $result_row->user_name; 
       $_SESSION['user_password'] = $result_row->user_password; 
       $_SESSION['user_logged_in'] = 1; 
     } else { 
      echo "Wrong Password"; 
     } 

}