Goodmorning everyone,
I'm developing a module for prestashop 1.7, at the moment I'm having problems intercepting the postprocess method in the main class of my module.
I need to do the checks on submit the form (which are on the user profile page, where I set personal information).
From what I understand, in a form a submit is made, the first thing that is called in a class is precisely the postProcess () method that takes care of validating the data received from the form just submissive (correct me if I'm wrong).
The problem is that when I submit my form it does not enter the postPorcess () method (I checked for a die ("test") and it does not even show the latter), while if I do the check I need by invoking my method staff inside a hook is made,
Can you tell me where I'm wrong?
Thank you very much and have a nice day.
Daniel.
Daniel,
This might be an endpoint problem, however, if you are sure to just handle the request via this Class, just use Tools::getValue('something_in_form') / Tools::isSubmit('var') to check that it's sent.
You don't really need to apply this one. If you need example, you should check Prestashop's native modules or Admin controllers, as it depends a lot of where you need to do this.
My thought after some years of module dev is that you should use a module front controller endpoint as you would with an API and a do a response in JSON like this example :
<?php
class DummyModuleNameAjaxModuleFrontController extends ModuleFrontController
{
public function initContent()
{
$response = array();
require_once _PS_MODULE_DIR_.'dummymodulename/dummymodulename.php';
$mod = new dummymodulename;
if (Tools::isSubmit('action') && Tools::isSubmit('var') && Tools::getValue('var') == $mod->getSomethingForSecurity()) {
$context = Context::getContext();
$cart = $context->cart;
switch (Tools::getValue('action')) {
case 'dummy_action_name':
// Don't forget to type it with an INT or secure this entry with strip_tags
$my_var = Tools::getValue('var');
break;
default:
break;
}
}
echo Tools::jsonEncode($response);
die;
}
}
Related
In Laravel 5.2, i need to do something while the authentication process is being triggered. (Not "after" the login.)
Here are what i have done.
In EventServiceProvider.php:
protected $listen = [
'Illuminate\Auth\Events\Attempting' => ['App\Listeners\UserLoginAttempt#handle'],
];
In app/Listeners/UserLoginAttempt.php:
<?php
namespace App\Listeners;
use Illuminate\Http\Request;
use Illuminate\Auth\Events\Login;
use Carbon\Carbon;
use Auth;
class UserLoginAttempt
{
public function __construct()
{
}
public function handle($event)
{
dd(Auth::user());
exit;
}
}
This always returns null.
** In fact, i have one "custom" field in the Login form, of which i need to capture from the above Event Handler. (I need to check something before actual Login event is done.)
How do i capture the Auth::user() from this (Illuminate\Auth\Events\Attempting) Handler please?
(In other words) How do i talk with the Login Form from the (Illuminate\Auth\Events\Attempting) Handler?
OR ----- Is the Auth::user() object created ONLY AFTER the login is fully processed? Which means i can not get it while in "attempting" stage?
Because i need to capture the login elements first, and if it is something "unauthorized" (after some calculations done in Handler), then i need to terminate the login attempt and route back to the login form from here.. by carrying the Failing Message along.
Please let me know what i'm missing here. Thank you.
According to documentation you are right that Auth::user() object is only created when user is authenticated.
Inside the Illuminate\Auth\Events\Attempting listener;
you can capture the form field by
public function handle(Attempting $event)
{
//return dd($event); // will echo all fields
$event->credentials['email']; // capturing the email field
}
you can also use the User model and then access the user table in DB, to get / check a user field in DB...
I want to change the the "views/layouts/main.php" to display the login form whenever the user isn't authenticated.
So I changed the "siteController" actionIndex like that:
public function actionIndex() {
$loginForm = new LoginForm();
$this->render('index', array('loginForm'=>$loginForm));
}
And then call it in "views/layouts/main.php" like that:
if(Yii::app()->user->isGuest):
echo $loginForm;
else :
echo 'JJJ';
endif;
Then when I go to my website, It display the error: "Undefined variable: loginForm".
I don't know how to fix this? :(
Define new property in your controller class:
public $loginForm;
In your main.php access it like:
echo $this->loginForm;
If you pass variable in your render method it will be available inside view file only, but not in layout file.
It's because the index template is loaded before main template. So, better way to do hat you want, is to define a public property in your Controller. I suggest you to define this property in Controller class because SiteController and *Controller extends it.
Then, you can run this.
if(Yii::app()->user->isGuest) {
echo $this->loginForm;
} else {
echo 'JJJ';
}
Pay attention, because in this way of work you go out MVC pattern. This way of work force you to define a LoginForm in each action. I suggest you to do that:
Leave clean your calls to render file.
public function actionIndex() {
$this->render('index');
}
And add a getLoginForm method in you Controller class obtaining:
if(Yii::app()->user->isGuest) {
echo $this->getLoginForm();
} else {
echo 'JJJ';
}
There are a couple issues here. Firstly, you are creating an object called $loginForm and assigning it a value of new LoginForm();
$loginForm = new LoginForm();
I'm not sure if you are doing this on purpose and LoginForm() is a function or a method that returns something, but I have a feeling you were intending to do:
$loginForm = new LoginForm;
Which creates a new instance of the class LoginForm (which is a default Yii webapp CFormModel class). Even if that is the case, there are better ways to do this.
The easiest way is to call a renderPartial of the already existing login.php view (located in protected/views/site/login.php) inside your index.php view like so:
if(Yii::app()->user->isGuest) {
$this->renderPartial("loginform",array("model"=>new LoginForm));
} else {
echo 'JJJ';
}
This renders the view login.php (without rendering the layout because we have already rendered the layout - here's the docs on render and renderPartial) and pass it a new instance of the model LoginForm assigned to a variable called $model.
You will most likely have to edit the look of login.php view to make it "fit", but keep in mind that this view is being used in the SiteController actionLogin as well.
All that's left to do then is modify your actionIndex to handle the form submission (you can just copy the existing SiteController actionLogin functionality)
Another nicer solution would be to create a widget for the login form which can be used all over your application. I'm not going to go into that, but you can read up about it here on SO or check out this tutorial or this one.
I'm constantly looking for the best way to use TDD in Yii app development. Nowday most web app are composed by a fronted, an API layer (usually JSON) to provide async calls to the server and a backend. By my side, most used test in this of app are unit tests and functional ones. The latter the most widely showed in guides and books leverage PHPUnit + Selenium, but Behat + Mink seems very cool too (but I'm not really confident with it yet)
If you ever used functional tests that use a browser (like Selenium) you know that the less you have to run them the better you feel. This cause they're slower, harder to maintain and sometimes (like the popup FB Login using JS SDK) are painful.
When working with a single web page application I care about testing JSON output of my apis. I'd like to test these functionalities with a unit test like approach in order to have faster tests that are easier to maintain. Considering that most of my Controller's action are availaible to Logged only user using accessControl filter I wondered on the best ways to have my tests up and running.
At this moment I think to have two ways to accomplish this
use cUrl toward the desired enpoint to get the JSON directly invoke
the controller's function
In the first scenario I can use fixtures but I got no way to mock CWebUser class (to emulate a logged user), using Apache when the cUrl comes it gets executed by an instance of my CWebApplication that is not the one executed by PHPUnit. I can get rid of this problem by making all my API calls stateless and, as a consequence, removing accessControl filter.
In the second one the only way I found to mock CWebUser class is to override it in the test class that I'm executing. This approach pays until I dont need to test use cases requiring different type of user, and I got no way to change at runtime (or at setup time) my webuser mock. The only way I found to mock my webuser is the one you can find below, this cause $this->getMock('WebUser') doesnt affect anyway CWebApplication's WebUser (read-only) singleton defined in the configuration file.
Here comes a concrete example:
class UserControllerTest extends CDbTestCase
{
public $fixtures=array(
/* NEEDED FIXTURES*/
);
public function testUserCanGetFavouriteStore() {
$controller = new UserController(1);
$result = json_decode($controller->actionAjaxGetFavStore());
$this->assertInternalType('array', $result->data);
$model = $result->data[0];
$this->assertEquals($model->name, "Nome dello Store");
}
}
class WebUser extends CWebUser {
public function getId() {
return 1;
}
public function getIsGuest() {
return false;
}
};
I was wondering if being able to authenticate with the api interface, either by an API key or a user/password combo could be useful.
This is ok if I move toward a almost stateless API integration, but most of the time I just have controller's actions (permitted to logged user only) that returns Json data to populate the frontend.
Anyone can suggest me a better method? Maybe it's just useless to test this kind of JSON output?
Best Regards
Maybe I'm oversimplifying your problem. It sounds like you want to emulate user logins before running the test? If that's the case, why not just create a User object in your fixture and actually log them in before running a test, and log them out after?
Something like:
/**
* Sets up before each test method runs.
* This mainly sets the base URL for the test application.
*/
protected function setUp()
{
parent::setUp();
// login as registered user
$loginForm = new UserLoginForm();
$loginForm->email = USER_EMAIL; // use your fixture
$loginForm->password = USER_PASSWORD; // use your fixture
if(!$loginForm->login()) {
throw new Exception("Could not login in setup");
}
}
protected function tearDown()
{
parent::tearDown();
Yii::app()->user->logout(true);
}
Ok actually the only solution that me and my team found is creating a stub WebUser class.
Rewriting WebUser class in this way you can authenticate a user without having Yii relying on the session.
class WebUserMock extends WebUser {
public function login($identity,$duration=0)
{
$id=$identity->getId();
$states=$identity->getPersistentStates();
if($this->beforeLogin($id,$states,false))
{
$this->changeIdentity($id,$identity->getName(),$states);
$duration = 0;
if($duration>0)
{
if($this->allowAutoLogin)
$this->saveToCookie($duration);
else
throw new CException(Yii::t('yii','{class}.allowAutoLogin must be set true in order to use cookie-based authentication.',
array('{class}'=>get_class($this))));
}
$this->afterLogin(false);
}
return !$this->getIsGuest();
}
public function changeIdentity($id,$name,$states)
{
$this->setId($id);
$this->setName($name);
$this->loadIdentityStates($states);
}
// Load user model.
protected function loadUser() {
$id = Yii::app()->user->id;
if ($id!==null)
$this->_model=User::model()->findByPk($id);
return $this->_model;
}
};
In the setUp method of your test class you can login any user (in this case leveraging my fixtures)
//a piece of your setUp() method....
$identity = new UserIdentity($this->users('User_2')->email, md5('demo'));
$identity->authenticate();
if($identity->errorCode===UserIdentity::ERROR_NONE)
Yii::app()->user->login($identity);
As a final thing to do just override the user component in the test configuration file and tell him to use this one:
protected/config/test.php
'user'=>array(
'class' => 'application.tests.mock.WebUserMock',
'allowAutoLogin'=>false,
),
Still not sure that this is the best way to handle it but seems to work fine
I have a “simple” question about the principle from the CodeIgniter MVC.
If I take a look in the manual from CI (Models) I see for example this:
function insert_entry()
{
$this->title = $_POST['title']; // please read the below note
$this->content = $_POST['content'];
$this->date = time();
$this->db->insert('entries', $this);
}
Well, ok – to put in data this way is bad I know :) but also if we user “$this->input->post()” … for me it doesn’t look better. Isn’t it better to handle the data in the controller before I use a function from a model? Maybe the model part looks so:
function insert_entry($data)
{
$this->db->insert('entries', $data);
}
And in the controller such like this:
$this->load->model('Blog');
$data = array();
$data['title'] = $this->input->post('title');
$data['content'] = $this->input->post('content');
$this->Blog->insert_entry($data);
But where i run the validation etc. … model or controller?
Maybe someone understand what I would like to know. Maybe you have some more experience, links or whatever. Thanks!
If you are trying to implement proper MVC or MVC-inspired design pattern with CodeIgniter, you have already failed. CodeIgniter does not follow the ideas of MVC and related patterns. It actually just clones the pattern used in Rails (I can elaborate in comments section, if you want to know why and how).
That said ...
The reason why $this->input->post() is used in controllers is to provide some abstraction and separate your code from PHP's superglobals. What you call a "controller" should collect data from the user's request and pass it to the model layer’s structures. The model layer should be completely unaware of the front-end. The domain business logic for creating an invoice does not change just because you renamed the <input/> for invoice number from "innr" to "number".
The data validation should happen in the model layer. When done properly, the code for validation is part of domain objects and data integrity checks would be handled by storage abstraction (for example, a data mapper), but in CodeIgniter people usually lump both domain and storage logic together and call it: "models". Of course that violated SRP, but CI users don't care and are even unaware of such principles. So basically, when writing for CI, the validation should happen in "models".
If you want to read more about the whole subject, you might find this post relevant.
hi you would have something like
class new_controller extends CI_Controller {
function __construct()
{
parent::__construct();
}
function insert_db_entry() {
$this->load->model('Blog');
$data = array();
if($this->input->post("submit")) {
$this->load->library("form_validation");
//create the form validation rules
if($this->form_validation->run() === TRUE) {
$data['title'] = $this->input->post('title');
$data['content'] = $this->input->post('content');
$this->Blog->insert_entry($data);
}
else {
$errors = validation_errors();
}
}
}
}
you use the form validation library to handle the validation when the form submit is detected.
I have a login button in the header of the website. This header's html is programmed into Zend framework views/layouts/home.phtml.
I have a hidden form in this layout that is triggered by jQuery thickbox inline content display integration. Reason, I dont want to make a ajax call to just fetch a small login form.
I create the form using Zend_Form and the problem is that I have to do it in all the controllers after checking if the user is logged in or not. I want to place this form generation in one single place, say in bootstrap and then have a logic in bootstrap to say that if user is logged in dont generate the form.
I don't know if bootstrap is the right place to do so or should I do it in some other place.
So, where should I instantiate the form so that its available everywhere if user is not logged in.
Create your own base controller which extends Zend_Controller_Action then have your controllers extend off of your base controller. I don't know what "jQuery thickbox inline content display integration" is...but you have several sections you can put it in depending when you need your code to run. init(), preDispatch(), postDispatch() etc... Just make sure when you extend off your base controller that you do sthing like:
parent::init()
parent::preDispatch()
parent::postDispatch()
etc... within each section so that the base code runs as well...
Be careful about Pradeep Sharma's solution (the answer he wrote himself and accepted below).
All the code code below is for ZF 1.12, and not ZF 2.0
In the bootstrap, Zend_Layout's MVC instance might not have been created yet. You should use Zend_Layout::startMvc() instead :
$view = Zend_Layout::startMvc()->getView() ;
And tbh I prefer executing this code in the preDispatch() function. New users of ZF might be interested in this :
application/plugins/HeaderForm.php :
class Application_Plugin_HeaderForm extends Zend_Controller_Plugin_Abstract
{
public function preDispatch(Zend_Controller_Request_Abstract $request)
{
$view = Zend_Layout::startMvc()->getView() ;
$view->headerForm = new Application_Form_HeaderForm() ;
}
}
Calling new Application_Form_HeaderForm() will autoload by default into application/forms/ folder. You can also create the form directly into the plugin with new Zend_Form(), and addElement() etc. but it won't be reusable.
Of course, you need to register this plugin in your bootstrap!
application/Bootstrap.php :
class Bootstrap extends Zend_Application_Bootstrap_Bootstrap
{
protected function _initPlugin()
{
$front = Zend_Controller_Front::getInstance() ;
$front->registerPlugin(new Application_Plugin_HeaderForm()) ;
}
}
Calling new Application_Plugin_HeaderForm() will autoload by default into application/plugins/ folder
I did it in a different way, extendingZend_Controller_Plugin_Abstract to implement a plugin and register it with front controller.
public function routeStartup(Zend_Controller_Request_Abstract $request) { }
generated the form inside the above mentioned method and by setting the form in $view object.
$view can be retrived using :
$view = Zend_Layout :: getMvcInstance()->getView();