Sorry, I'm a beginner at Symfony and I've tried to find an answer but nothing worked. I'm using Symfony3.4 which was updated from Symfony 2.8 a few months ago.
Now, I'm trying to do a rather simple thing : using a formType in a Controller, but no matter what, Symfony keeps showing the following error : Could not load type "Cha\GeneralBundle\Form\StripePaymentType": class does not exist.
Here's my StripePaymentType -really, really basic:
namespace Cha\GeneralBundle\Form;
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\Extension\Core\Type\TextType;
use Symfony\Component\Form\FormBuilderInterface;
class StripePaymentType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder->add('name', TextType::class);
}
public function getBlockPrefix()
{
return 'cha_stripe_payment_form';
}
}
Here's my Controller action, once again, a basic thing (I didn't even write any code in there, because of this strange error) :
public function upgradeOfferPaymentAction(Offers $offers)
{
$form = $this->createForm(StripePaymentType::class);
return $this->render(
'#ChaGeneral/Offers/offer_payment.html.twig', array(
'form' => $form->createView()
));
}
I tried to use my form as a service but it did not work either :
cha.stripe.payment.form:
class: Cha\GeneralBundle\Form\StripePaymentType
tags:
- { name: cha_stripe_payment_form }
I'm probably missing something but I can't figure what...
Thank in advance for your help!
You must add this line at the top of the form class:
use Symfony\Component\Form\Extension\Core\Type\TextType;
Revert the SF version is not a solution, is just dodge the problem.
Well, finally I found what caused this problem : somebody did a composer update and we updated from Symfony 3.4.13 to Symfony 3.4.14. A simple revert did the trick, now everything is working again.
Thanks a lot for your help!
I installed a brand new, clean Joomla 3.4.5 and then installed a component written by myself, which worked totally fine in Joomla 2.5. In Joomla 3 however, I get server error 500... in some cases...
I narrowed the error down to the following weird situation:
The component is called com_confighdv (I'm extending Joomla core's com_config). I added a view called JustaName, existing of two files:
admin/views/justaname/view.html.php:
<?php
class ConfigHdVViewJustaName extends JViewLegacy
{
}
?>
admin/views/justaname/tmpl/default.php:
Hello world!
This works fine when I go to index.php?option=com_confighdv&view=justaname.
Then I change the view's name from JustaName to Component:
The view's folder becomes: admin/views/component/
Class declaration becomes: class ConfigHdVViewComponent extends JViewLegacy {}
Now, when I go to index.php?option=com_confighdv&view=component I get a server error 500 :s
I really don't know what to do with this. Help is very much appreciated!
Solved! Switching to Joomla's maximum error reporting provided the explanation:
Fatal error: Class ConfigHdVModelComponent contains 1 abstract method and must therefore be declared abstract or implement the remaining methods (JModelForm::getForm) in /xxx/administrator/components/com_confighdv/models/component.php on line 18
So, the problem is not in the view, but in the model that belongs to the Component view!
I accidentally created this problem myself, by reducing the model declaration to what I thought was the absolute minimum:
class ConfigHdVModelComponent extends JModelAdmin
{
{
While this is not allowed, because you always have to define the getForm method, so:
class ConfigHdVModelComponent extends JModelAdmin
{
public function getForm($data = array(), $loadData = true)
{
}
{
I want to create a maintenance Page for my cake website by checking a Database Table for a maintenance flag using a sub-function of my AppController "initilize()" method. If the flag is set, i throw my custom MaintenanceException(Currently containing nothing special):
class MaintenanceException extends Exception{
}
To handle it, I implemented a custom App Exception Renderer:
class AppExceptionRenderer extends ExceptionRenderer {
public function maintenance($error)
{
return "MAINTENANCE";
}
}
I am able to see this maintenance Text on my website if I set my DB flag to true, but I could not find any information in cake's error handling documentation (http://book.cakephp.org/3.0/en/development/errors.html) on how I can actually tell the Exception renderer to render view "maintenance" with Template "infopage".
Can I even us that function using the ExceptionRenderer without a custom error controller? And If not, how should a proper ErrorController implementation look like? I already tried this:
class AppExceptionRenderer extends ExceptionRenderer {
protected function _getController(){
return new ErrorController();
}
public function maintenance($error)
{
return $this->_getController()->maintenanceAction();
}
}
together with:
class ErrorController extends Controller {
public function __construct($request = null, $response = null) {
parent::__construct($request, $response);
if (count(Router::extensions()) &&
!isset($this->RequestHandler)
) {
$this->loadComponent('RequestHandler');
}
$eventManager = $this->eventManager();
if (isset($this->Auth)) {
$eventManager->detach($this->Auth);
}
if (isset($this->Security)) {
$eventManager->detach($this->Security);
}
$this->viewPath = 'Error';
}
public function maintenanceAction(){
return $this->render('maintenance','infopage');
}
}
But this only throws NullPointerExceptions and a fatal error. I am really dissapointed by the cake manual as well, because the code examples there are nowhere close to give me an impression of how anything could be done and what functionality I actually have.
Because I had some more time today, I spent an hour digging into the cake Source and found a solution that works well for me (and is propably the way it should be done, altough the cake documentation does not really give a hint):
Step 1: Override the _template(...)-Method of the ExceptionRenderer in your own class. In my case, I copied the Method of the parent and added the following Code at the beginning of the method:
$isMaintenanceException = $exception instanceof MaintenanceException;
if($isMaintenanceException){
$template = 'maintenance';
return $this->template = $template;
}
This tells our Renderer, that the error Template called "maintentance"(which should be located in Folder: /Error) is the Error Page content it should render.
Step 2: The only thing we have to do now (And its is kinda hacky in my opinion, but proposed by the cake documentation in this exact way) is to set the layout param in our template to the name of the base layout we want to render with. So just add the following code on top of your error template:
$this->layout = "infopage";
The error controller I created is actually not even needed with this approach, and I still don't know how the cake error controller actually works. maybe I will dig into this if I have more time, but for the moment.
Error like:The view 'LoginRegister' or its master was not found or no view engine supports the searched locations. The following locations were searched:
~/Views/MyAccount/LoginRegister.aspx
~/Views/MyAccount/LoginRegister.ascx
~/Views/Shared/LoginRegister.aspx
~/Views/Shared/LoginRegister.ascx
~/Views/MyAccount/LoginRegister.cshtml
~/Views/MyAccount/LoginRegister.vbhtml
~/Views/Shared/LoginRegister.cshtml
~/Views/Shared/LoginRegister.vbhtml
Actually my page view page is ~/Views/home/LoginRegister.cshtml so what i do
and my RouteConfig is
public class RouteConfig
{
public static void RegisterRoutes(RouteCollection routes)
{
routes.IgnoreRoute("{resource}.axd/{*pathInfo}");
routes.MapRoute(
name: "Default",
url: "{controller}/{action}/{id}",
defaults: new { controller = "MyAccount", action = "LoginRegister", id = UrlParameter.Optional }
);
}
}
Be careful if your model type is String because the second parameter of View(string, string) is masterName, not model. You may need to call the overload with object(model) as the second parameter:
Not correct :
protected ActionResult ShowMessageResult(string msg)
{
return View("Message",msg);
}
Correct :
protected ActionResult ShowMessageResult(string msg)
{
return View("Message",(object)msg);
}
OR (provided by bradlis7):
protected ActionResult ShowMessageResult(string msg)
{
return View("Message",model:msg);
}
Problem:
Your View cannot be found in default locations.
Explanation:
Views should be in the same folder named as the Controller or in the Shared folder.
Solution:
Either move your View to the MyAccount folder or create a HomeController.
Alternatives:
If you don't want to move your View or create a new Controller you can check at this link.
In Microsoft ASP.net MVC, the routing engine, which is used to parse incoming and outgoing URL Combinations, is designed with the idea of Convention over Configuration. What this means is that if you follow the Convention (rules) that the routing engine uses, you don't have to change the Configuration.
The routing engine for ASP.net MVC does not serve web pages (.cshtml). It provides a way for a URL to be handled by a Class in your code, which can render text/html to the output stream, or parse and serve the .cshtml files in a consistent manner using Convention.
The Convention which is used for routing is to match a Controller to a Class with a name similar to ControllerNameController i.e. controller="MyAccount" means find class named MyAccountController. Next comes the action, which is mapped to a function within the Controller Class, which usually returns an ActionResult. i.e. action="LoginRegister" will look for a function public ActionResult LoginRegister(){} in the controller's class. This function may return a View() which would be by Convention named LoginRegister.cshtml and would be stored in the /Views/MyAccount/ folder.
To summarize, you would have the following code:
/Controllers/MyAccountController.cs:
public class MyAccountController : Controller
{
public ActionResult LoginRegister()
{
return View();
}
}
/Views/MyAccount/LoginRegister.cshtml: Your view file.
In your LoginRegister action when returning the view, do below, i know this can be done in mvc 5, im not sure if in mvc 4 also.
public ActionResult Index()
{
return View("~/Views/home/LoginRegister.cshtml");
}
Check the build action of your view (.cshtml file) It should be set to content. In some cases, I have seen that the build action was set to None (by mistake) and this particular view was not deploy on the target machine even though you see that view present in visual studio project file under valid folder
This could be a permissions issue.
I had the same issue recently. As a test, I created a simple hello.html page. When I tried loading it, I got an error message regarding permissions. Once I fixed the permissions issue in the root web folder, both the html page and the MVC rendering issues were resolved.
Check whether the View (.ASPX File) that you have created is having the same name as mentioned in the Controller. For e.g:
public ActionResult GetView()
{
return View("MyView");
}
In this case, the aspx file should be having the name MyView.aspx instead of GetView.aspx
I got this error because I renamed my View (and POST action).
Finally I found that I forgot to rename BOTH GET and POST actions to new name.
Solution : Rename both GET and POST actions to match the View name.
If the problem happens intermittently in production, it could be due to an action method getting interrupted. For example, during a POST operation involving a large file upload, the user closes the browser window before the upload completes. In this case, the action method may throw a null reference exception resulting from a null model or view object. A solution would be to wrap the method body in a try/catch and return null. Like this:
[HttpPost]
public ActionResult Post(...)
{
try
{
...
}
catch (NullReferenceException ex) // could happen if POST is interrupted
{
// perhaps log a warning here
return null;
}
return View(model);
}
I had this same issue.
I had copied a view "Movie" and renamed it "Customer" accordingly.
I also did the same with the models and the controllers.
The resolution was this...I rename the Customer View to Customer1 and
just created a new view and called it Customer....I then just copied
the Customer1 code into Customer.
This worked.
I would love to know the real cause of the problem.
UPDATE
Just for grins....I went back and replicated all the renaming scenario again...and did not get any errors.
I came across this error due to the improper closing of the statement,
#using (Html.BeginForm("DeleteSelected", "Employee", FormMethod.Post))
{
} //This curly bracket needed to be closed at the end.
In Index.cshtml view file.I didn't close the statement at the end of the program. instead, I ended up closing improperly and ran into this error.
I was sure there isn't a need of checking Controller ActionMethod code because I have returned the Controller method properly to the View. So It has to be the view that's not responding and met with similar Error.
If you've checked all the things from the above answers (which are common mistakes) and you're sure that your view is at the location in the exceptions, then you may need to restart Visual Studio.
:(
In my case, I needed to use RedirectToAction to solve the problem.
[HttpGet]
[ControleDeAcessoAuthorize("Report/ExportToPDF")]
public ActionResult ExportToPDF(int id, string month, string output)
{
try
{
// Validate
if (output != "PDF")
{
throw new Exception("Invalid output.");
}
else
{
...// code to generate report in PDF format
}
}
catch (Exception ex)
{
return RedirectToAction("Error");
}
}
[ControleDeAcessoAuthorize("Report/Error")]
public ActionResult Error()
{
return View();
}
I ran into this a while ago and it drove me crazy because it turned out to be simple. So within my View I was using a grid control that obtained data for the grid via an http request. Once the middle tier completed my request and returned the dataset, I received the same error. Turns out my return statement was 'return View(dataset);' instead of 'return Json(dataset);
I couldn't find any solution to this problem, until I found out the files didn't exist!
This took me a long time to figure out, because the Solution Explorer shows the files!
But when I click on Index.cshtml I get this error:
So that was the reason for this error to show. I hope this answer helps somebody.
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.