PHP - Moved: Model In Controller
This topic has been moved to Application Frameworks.
http://www.phpfreaks.com/forums/index.php?topic=318251.0 Similar TutorialsThis topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=306553.0 Moving an application from one server and domain to another. Have updated all references that I can find to the old url and db. Mirrored all permissions, etc. Am getting the following notice: Notice (8): Undefined index: CakeCookie [APP/config/bootstrap.php, line 47] $user_preferred_city = ''; // Checking whether user has already visited our site $user_preferred_city = $_COOKIE['CakeCookie']['signup_city_name']; $boot = true $shellPaths = null $localePaths = null $vendorPaths = null $pluginPaths = null $helperPaths = null $viewPaths = null $componentPaths = null $controllerPaths = null $behaviorPaths = null $modelPaths = null $user_preferred_city = "" $user_preferred_city = ''; // Checking whether user has already visited our site $user_preferred_city = $_COOKIE['CakeCookie']['signup_city_name']; include - APP/config/bootstrap.php, line 47 Configu :__loadBootstrap() - CORE/cake/libs/configure.php, line 660 Configu :getInstance() - CORE/cake/libs/configure.php, line 137 include - CORE/cake/bootstrap.php, line 43 [main] - APP/webroot/index.php, line 81 What should I look for to resolve this? Am new to PHP and more of a dba than anything, so I a really appreciate any help you can offer. Thanks! This topic has been moved to Application Frameworks. http://www.phpfreaks.com/forums/index.php?topic=353375.0 This topic has been moved to mod_rewrite. http://www.phpfreaks.com/forums/index.php?topic=325383.0 This topic has been moved to Application Frameworks. http://www.phpfreaks.com/forums/index.php?topic=355229.0 hi everyone I have this site i want to make changes to the contact form. Right now it sends the info to a database and posts in the admin area. I have this file controller.php that has a section with a funtion function Cuenta_Contactos that i think passes the info. Is there a way to add an email address there so it sends the message to that email account? Thanks ive attached the file . I'm using this struture to run my application, everything works fine, but i don't like how the controllers are called. If i load the model in construct it will work great again, but i think that is unnecessary. You can see two pieces of code about controller "Post.php", the last one, works great. If i try this: (Posts.php) namespace app\controllers; use app\libraries\Controller; use app\models\User; class Posts extends Controller { public function index():void { //Get posts $user = new User(); $posts = $user->posts(); $data = [ 'posts' => $posts ]; $this->view('posts/index', $data); } } I will get an error: QuoteFatal error: Uncaught Error: Class 'app\models\User' not found in C:\xampp\htdocs\shake\app\controllers\Posts.php Structu Quote
- app bootstrap.php //include all files from libraries require_once __DIR__ . "/../vendor/autoload.php"; index.php use app\libraries\Core; require_once '../app/bootstrap.php'; //Init Core Library $init = new Core(); core.php public function __construct(){ $url = $this->getUrl(); //Look in controllers for first value if(file_exists(dirname(__FILE__, 2) . '/controllers/' . ucwords($url[0]) . '.php')){ //if exists, set as controller $this->currentController = ucwords($url[0]); //Unset 0 index unset($url[0]); } //Require the controller require_once dirname(__FILE__, 2) . '/controllers/' . $this->currentController . '.php'; //instantiate controller class $class = 'app\\controllers\\' . $this->currentController; $this->currentController = new $class; //check for second part of url if(isset($url[1])){ //Check to see if method exists in controller if(method_exists($this->currentController, $url[1])){ $this->currentMethod = $url[1]; //Unset 1 index unset($url[1]); } } //Get params $this->params = $url ? array_values($url) : []; //Call a callback with array of params call_user_func_array([$this->currentController, $this->currentMethod], $this->params); } Posts.php class Posts extends Controller { public function __construct() { $this->userModel = $this->model('User'); $this->postModel = $this->model('Post'); } public function index():void { //Get posts $posts = $this->userModel->posts(); $data = [ 'posts' => $posts ]; $this->view('posts/index', $data); } } Controller.php abstract class Controller { /** * Load model * @param $model * @return mixed */ public function model($model) { //Require model file require_once '../app/models/' . $model . ".php"; //instantiate model $class = 'app\models\\' . $model; return new $class(); }
I'm very new to MVC and I'm trying to figure out how to transition my code and I'm not sure where to place things. It easy to want to just put everything in the view, but I know that doesn't make sense. Can you please guide me on how I should transition the rest of my code for my header? I'm currently using codeigniter. Current View Code: [Select] <?php $page = substr(end(explode(DIRECTORY_SEPARATOR, $_SERVER['PHP_SELF'])), 0, -4); $title = (array_key_exists($page, $page_names) !== false) ? $page_names[$page]: ''; if (array_key_exists($page, $page_names) !== false) { $title .= " | Jason Biondo"; } $banner_imgs = array('contact.jpg', 'about.jpg', 'tools.jpg', 'portfolio.jpg', 'articles.jpg'); $nav_names = array('Contact', 'About', 'Tools', 'Portfolio', 'Articles'); $gutter_values = array('307', '230', '161', '84', '0'); $alt_page = 'Articles'; foreach($nav_names as $k => $name) { if($page === strtolower($name)) { $g_value = $gutter_values[$k]; } } if(!isset($g_value)) { $g_value = 0; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" lang="en" dir="ltr"> <head> <meta http-equiv="content-type" content="text/html;charset=utf-8" /> <meta http-equiv="content-script-type" content="text/javascript" /> <meta http-equiv="content-style-type" content="text/css" /> <meta http-equiv="content-language" content="en" /> <meta name="description" content="Jason is a serial entrepreneur that builds web applications using advanced programming technologies and unique interface design." /> <meta name="keywords" content="jason , entrepreneur, investors, venture capitalist, angel investor, vc, ventures, private equity, startups, startup community, startup investments, investment network, raise capital, where to find capital, fund raising, venture financing, contact investors, angel fund, angel group, investment strategy, business plan" /> <title><?=$title?></title> <link rel="shortcut icon" type="image/x-icon" href="/assets/img/favicon.ico" /> <link rel="stylesheet" type="text/css" media="screen" href="/css/style.css" /> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.5.1/jquery.min.js"></script> <script src="./js/jquery.validate.min.js"></script> <script src="./js/jquery.corner.js"></script> <script src="./js/init.js"></script> <?php if (file_exists("./assets/js/pages/${page}.js")) { echo "<script type=\"text/javascript\" src=\"./js/pages/${page}.js\"></script>"; } ?> </head> <body> <div class="header"> <div class="header_container"> <a href="/"> <img class="logo" src="/assets/img/logo.png" alt="Jason" title="Jason" /> </a> <div class="navigation f_right"> <div class="gutter"> <span id="nav_highlight" class="nav_highlight" style="left: <?php echo $g_value; ?>px;"></span> </div> <ul> <?php foreach($nav_names as $k => $name) { if(is_odd($k)) { ?> <li class="pike"></li> <?php } ?> <li> <a id="<?php echo strtolower($name); ?>" <?php echo (($page === strtolower($name)) || ($alt_page === $name)) ? 'class="selected"' : ''; ?> href="/<?php echo ($name === $alt_page) ? '' : strtolower($name); ?>"><?php echo $name; ?></a> </li> <?php if(is_odd($k)) { ?> <li class="pike"></li> <?php } } ?> </ul> </div> </div> </div> <div class="stripe"></div> <div id="content_container" class="content_container"> <div id="banner" class="banner"> <?php foreach($nav_names as $k => $name) { if(($page === strtolower($name)) || ($alt_page === $name)) { ?> <img src="/assets/img/banners/<?php echo $banner_imgs[$k]; ?>" /> <h1><?php echo $name; ?></h1> <?php break; } if($alt_page === $name) { ?> <img src="/assets/img/banners/<?php echo $banner_imgs[$k]; ?>" /> <h1><?php echo $name; ?></h1> <?php } } ?> </div> <div class="banner_stripe"></div> <div class="content"> Current Controller Code: [Select] <?php if ( ! defined('BASEPATH')) exit('No direct script access allowed'); class Writing extends CI_Controller { public function index() { $this->load->library('common'); $data['page_names'] = $this->common->page_names(); $data['is_odd'] = $this->common->is_odd(); $this->load->view('includes/header', $data); $this->load->view('writing_view'); $this->load->view('includes/footer'); $this->output->enable_profiler(TRUE); } } Hi I will try and explain this as best I can, as I am not sure if the "include" is causing the problem. What I am trying to achieve is count the total number of vehicles in the "vehicles" db table and pass it to views (I have recently moved over to MVC so still learning so please be kind and keep it simple for me please) I have a folder in views called "vehicletracker" , within this folder I have 2 files "index.php" and "widget.php".........The "widget.php" file is an "include" within "index php". The problem I am having is passing data from the controller to the view, I keep getting an "Undefined index: total" error and wonder if someone can help and show me where I am going wrong. This is my Model
<?php class Widget { private $db; public function __construct(){ $this->db = new Database; } public function countVehicles(){ $this->db->query("SELECT * FROM vehicles"); return $this->db->rowCount(); } } This is my controller
<?php class Widgets extends Controller{ public function __construct(){ $this->widgetModel = $this->model('Widget'); public function widget (){ $data['total'] = $this->widgetModel->countVehicles(); $this->view('vehicletracker/widget', $data); } } And my view
<h2 class="text-white"><?php echo $data['total']; ?></h2>
Thanks in advance for any help you can give I am currently building my own MVC Framework and I have run into an issue that I can't solve when attempting to use 2 methods from a single model. I know that this isn't an issue with the queries or information being return. I am unsure of the proper way that I can call 2 methods when checking to see if a user is logged in... The issue that I face is both methods are calling the DB and it throws a PDO error which I don't know how to get around the issue.. Any guidance would be nice as I have been banging my head over this issue. Thank you!
Fatal error: Uncaught Error: Call to undefined method PDO::Select() in C:\xampp\htdocs\PicWrist\app\models\user.class.php:318 Stack trace: #0 C:\xampp\htdocs\PicWrist\app\controllers\home.php(21): User->getUser('4') #1 C:\xampp\htdocs\PicWrist\app\core\app.php(36): Home->index() #2 C:\xampp\htdocs\PicWrist\app\init.php(47): App->__construct() #3 C:\xampp\htdocs\PicWrist\public\index.php(5): require('C:\\xampp\\htdocs...') #4 {main} thrown in C:\xampp\htdocs\PicWrist\app\models\user.class.php on line 318
//Controller <?php /** * Load the View */ class Controller{ public function view($view, $data = []) { if (file_exists('../app/views/' . $view . '.php')) { require_once '../app/views/' . $view . '.php'; } } public function model($model, $data = []) { if (file_exists('../app/models/' . $model . '.class.php')) { require_once '../app/models/' . $model . '.class.php'; return new $model(); } else return false; } } ?> //Home Controller <?php Class Home extends Controller { public static $user; public $errors; public function __construct() { self::$user = $this->model('user'); } public function index() { $userdata = []; $data = []; show(self::$user); $isLoggedin = self::$user->isLoggedin; show($isLoggedin); $userdata = self::$user->getUser($_SESSION['user_id']); show($userdata); $this->view('home', $data); } } ?> <?php /** * Users */ class User { private $db; private $errors = ''; public $isLoggedin = False; public $isAdmin = False; public function __construct() { $this->isLoggedIn(); } public function isLoggedIn() { if(isset($_SESSION['user_id'])){ $data['user_id'] = $_SESSION['user_id']; $db = Database::getInstance(); $query = "SELECT * FROM users WHERE user_id = :user_id LIMIT 1"; $results = $db->Select($query, $data); if (is_array($results)) { $this->isLoggedIn = True; } } } public function getUser($id) { if(isset($id)) { $data['user_id'] = intval($id); $db = Database::getInstance(); $query = 'SELECT * FROM users WHERE user_id = :user_id'; $results = $db->Select($query, $data); if (is_array($results)) { return $results; } else { return False; } } else { return False; } } //DB <?php /** * Database Connection */ class Database { private $dbHost = DB_HOST; private $dbUser = DB_USER; private $dbPass = DB_PASS; private $dbName = DB_NAME; private $statment; private static $dbHandler; private $error; public function __construct() { $conn = 'mysql:host=' . $this->dbHost . ';dbname=' . $this->dbName; $options = array( PDO::ATTR_PERSISTENT => true, PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION ); try { self::$dbHandler = new PDO($conn, $this->dbUser, $this->dbPass, $options); } catch (PDOException $e) { self::$error = $e->getMessage(); echo self::$error; } } public static function getInstance() { if(self::$dbHandler) { return self::$dbHandler; } return $instance = new Self(); } public function Select($query, $data = array()) { $statement = self::$dbHandler->prepare($query); $result = $statement->execute($data); If($result) { $data = $statement->fetchAll(PDO::FETCH_OBJ); if(is_array($data)) { // show($data); return $data; } } return False; } public function Update($query, $data = array()) { $statement = self::$dbHandler->prepare($query); $result = $statement->execute($data); If($result) { return True; } return False; } } ?> Edited July 14 by avargas94 Wrong Code Hi;
I have succesfully managed to route all the traffic to the file controller.php using this command line of .htaccess :
RewriteEngine On RewriteRule (.*) /var/www/html/site/controller.php [L,QSA]Now how to tell the controller.php to process the url that was passed. For example i have tried that: controller.php : <?php include( $_SERVER["REQUEST_URI"] ); ?>but the controller.php didn't do anything i keep getting a blank page when i browse any webpage such as site/index.php so how to tell controller.php to echo the content of the request ? because the user shouldn't notice any change as if there is no controller.php. Thank you. I have a list of checkboxes, like e.g each checkbox has a value as ResumeID, now question is/are; 1) how to get the values of the checkboxes using jQuery ? 2) when I got the values of the checkboxes using jQuery, how to pass those values to a PHP Controller action ? I am developing a new promising service (media-industry) in the Los Angeles area. Contact me if interested.
Leroy. liroyleshed@gmail.com Hi, I have a cakePHP program that works and displays data from a database table. The part I dont understand and I couldnt find anything on the cakephp blog as yet was the model was left blank? q1) The model is left blank but still works so what is the default behaviour here for the model? It doesnt seem to be needed to do anything i?. q2)I am selecting all the data from 1 table but how do I add a sql statement to select part of the table or a join of 2 tables? the find 'all' seems to just do that only in finding all data from a table. <?php class Post extends AppModel { /*var $name='User';*/ } ?> <?php class PostsController extends AppController { public $helpers=array('Html','Form'); public function index() { $this->set('posts',$this->Post->find('all')); } public function home() { $this->set('posts',$this->Post->find('all')); } } ?> view /// .. <?php foreach ($posts as $item): echo '<tr><td>'. $item['Post']['id']. '</td>'; echo '<td>'. $item['Post']['title'].'</td>'; echo '<td>'. $item['Post']['body'].'</td>'; echo '<td>'. $item['Post']['created'] .'</td>'; echo '</tr>'; endforeach; unset($item); ?> Hello I'm developing a Model-View-Controller framework for my personal learning curve. I've created an active record class and I've got stuck when using variables in a where clause. My problem is this: When, in a where statement, you want to find the Users with the first name Mike, you might write SELECT Firstname, Surname FROM Users WHERE Firstname = 'Mike' Where Mike is enclosed with quotation marks. When you're comparing entities quotation marks shouldn't be used. For example, SELECT Firstname, Surname, OrderTitle FROM Users, Orders WHERE Firstname = 'Mike' AND Users.Firstname = Orders.Firstname So when using variables in my where statements, like $Name = Mike $this->Where(Firstname',$Name); $this->Where('Users.Firstname','Orders.Firstname'); How do I differentiate, in my function Where() when to enclose the string with quotation marks, i.e when it is a (Table.Entity) rather than a constant. Thank you Can a person instantiate a model? If so how do you do it. I already know how to instantiate a class thank you Hey all,
I am making a application that requires a user to login in. I have a user model, which does such things as finds, registers, logins in, updates etc a user. It also grabs all of a users data from a database, such as their username, email address etc and checks whether or not a user is logged in.
The problem I face is this. I want to control what content a user can see depending on if they are logged in or not. In addition to this, I want to output the users data, such as their username, in other parts of the website – such as the header and footer. There is no real easy way of doing this, and requires the user model to be instantiated on every page. My question is should this class be a model or just a core library class?
Hi Everyone. I am in need of assistance, I need a form that i will use on a model agency site i am making, the form i have only has the option to upload one image and records name, age etc... What i need is a form that will upload multiple images, 2 will do but if i had a option to add more that would be great, and a pdf resume, does anyone have a similar form i can have a look at that does this, any help be awesome, i am already over the deadline. Hi all,
I was just wondering is this possible what I'm trying to do?
Models can only store one object at a time...
I'm trying to access multiple objects via one model...
<?php class users extends Model { protected $username; protected $firstname; protected $secondname; protected $age; // There will be more here... public function __construct($id = NULL) { if($id != NULL) { $this->username = $sql->getUsername($id); $this->firstname = $sql->getFirstname($id); $this->secondname = $sql->getSecondname($id); $this->age = $sql->getAge($id); } } public function numOfUsers() { $num = $sql->countAllUsers($query); return $num; } public function getUsername() { return $this->username; } public function getFirstname() { return $this->firstname; } public function getSecondname() { return $this->secondname; } public function getAge() { return $this->age; } }; class UsersView extends View { $users = new users(); for($i = 0; $i < $users->numOfUsers(); $i++) { $user = new users($i); echo "Username: {$user->getUsername()}"; echo "Username: {$user->getFirstname()}"; echo "Username: {$user->getSecondname()}"; echo "Username: {$user->getAge()}"; } }; ?> Dear Members, I am a PHP developer and desisgning a WAP page for mobiles.I need some php code to find out the model of a mobile phone from where request is coming so that based on that information I can upload the proper CSS for my page.If you can give me some hints as well that would also help me to proceed further.Currently I am able to fetch user-agent of mobile but not able to find out the specific model of mobile. |