I have created an rest calling framework in YII2. In this I have a class APIRequest From this class I want to render a page if I get an error from API.
My code:
public static function response($response,$serviceObject)
{
if($serviceObject->responseCode == 420)
{
$errorMessage = $response->errorMessage;
return \Yii::$app->getView()->renderFile('#app/views/merchants/error.php',['errorMessage'=>$errorMessage]);
}
else
{
return $response;
}
}
But this is not working.
Not much information to go on. where is this method invoked? in the controller class?
perhapse this could help http://www.yiiframework.com/doc-2.0/guide-runtime-handling-errors.html
Generally the Yii REST API doesn't use any views. See
http://www.yiiframework.com/doc-2.0/guide-rest-quick-start.html
use
return \Yii::$app->view->renderFile('#app/views/merchants/error.php',['errorMessage'=>$errorMessage]);
instead of
return \Yii::$app->getView()->renderFile('#app/views/merchants/error.php',['errorMessage'=>$errorMessage])
Related
I have this requirements. I need to be able to write this code in my razor views:
#Filters.Render(Filters.DateRangeFilter, new DateRangeFilterParameters { });
The alternative is:
#Html.Partial("/Views/Shared/DateRangeFilter.cshtml", new DateRangeFilterParameters { });
In other words, I want Filters class to wrap Html.Partial. For that reason, I thought of this code:
public class Filters {
public const string DateRangeFilter = "/Views/Shared/DateRangeFilter.cshtml";
public static HtmlString Render(string filterPath, object parameters)
{
// Here I need to call Html.Partail, how?
}
}
To use Html.Raw within the controller you can request the injected IHtmlHelper service. E.g.:
HttpContext.RequestServices.GetService(typeof(IHtmlHelper)) as IHtmlHelper;
Or you can do your own implementation for the helper. And in order to use Html.Partial you need to use IRazorViewEngine, ViewContext, and other stuff. So basically you need to implement a service for that, and here is a good example Render Partial View To String Outside Controller Context.
I don't know if there is an easier way to achieve those, but that is what on my mind.
I'm trying to call an action on a controller in an MVC project from a view and I get the following error:
This can happen when a controller uses RouteAttribute for routing, but no action on that controller matches the request
I've read some people have removed the attribute routing to get this to work but that seems a bit extreme. Does anyone know where to start with this one?
//Calling in view like so
#Html.Action("Edit", new { datablockId = 227 })
//THe controller
[RoutePrefix("CustomData")]
public class CustomDataController : Controller, ICustomDataController
{
[Route("Edit")]
[HttpGet]
public ActionResult Edit(int datablockId)
{
return this.PartialView(new CustomDataEditViewModel() { DataRows = Data, DataBlockId = datablockId });
}
}
Try routing the action to that particular controller explicitly like this:
#Html.Action("Edit", "CustomData" ,new { datablockId = 227 })
Html action accepts aditional parameters that might fix your routing issue, those parameters are: Html.Action("Action", "Controller", Parameters)
How can I get current action?
This code:
if (!Yii::$app->controller->action->id == 'lang') {
Url::remember();
}
returns an error:
PHP Notice – yii\base\ErrorException
Trying to get property of non-object
You should use beforeAction() event instead of init().
Also you can simply use $this because it contains current controller.
public function beforeAction($action)
{
if (parent::beforeAction($action)) {
if ($this->action->id == 'lang') {
Url::remember();
}
return true; // or false if needed
} else {
return false;
}
}
if use Yii2 in view - try this:
$this->context->action->id;
You can get current action name by:
Yii::$app->controller->action->id
And get the controller name with this one:
Yii::$app->controller->id;
Note: Remember that these will only work after the application has been initialized. Possible use: inside a controller action/ inside a model or inside a view
Reference: Class yii\web\Controller
You can get current action id by :)
Yii::$app->controller->id;
I made a file in library/My/Utils/Utils.php. The content of the file is :
class My_Utils_Utils{
public function test(){
$this->_redirect('login');
}
}
This class is called from a layout; the problem is with the _redirect(); I get this error : The page isn't redirecting properly. My question is how call the _redirect() function from a class made by you in ZEND framework 1 .
Thanks in advance.
Use redirect() instead of _redirect(). Usage is:
$this->redirect(<action>, <controller>, <module>, <param>);
In your case $this->redirect('login'); should do the trick.
You can use the redirector action-helper, which you can get statically from the HelperBroker using:
// get the helper
$redirectHelper = Zend_Controller_Action_HelperBroker::getStaticHelper('redirector');
// call methods on the helper
$redirect->gotoUrl('/some/url');
It should be noted, however, that the layout is considered part of the view layer. Typically, any checks that result in a redirect should probably take place earlier in the request dispatch-cycle, typically in a controller or in a front-controller plugin.
The _redirect function is provided by the class Zend_Controller_Action. You can fix this in two ways :
Extend Zend_Controller_Action and use _redirect
class My_Utils_Utils extends Zend_Controller_Action {
public function test(){
$this->_redirect('login');
}
}
in layout:
$request = Zend_Controller_Front::getInstance()->getRequest();
$response = Zend_Controller_Front::getInstance()->getResponse()
$util = new My_Utils_Utils($request, $response); // The constructor for Zend_Controller_Action required request and response params.
$util->test();
Use gotoUrl() function Zend_Controller_Action_Helper_Redirector::gotoUrl()
$redirector = new Zend_Controller_Action_Helper_Redirector();
$redirector->gotoUrl('login');
//in layout :
$util = new My_Utils_Utils();
$util->test();
I am new to Zend framework and I have a problem.
I created a controller abstract class which implements the functions like:
protected function AddError($message) {
$flashMessenger = $this->_helper->FlashMessenger;
$flashMessenger->setNamespace('Errors');
$flashMessenger->addMessage($message);
$this->view->Errors = $flashMessenger->getMessages();
}
protected function activateErrors()
{
$flashMessenger = $this->_helper->FlashMessenger;
$flashMessenger->setNamespace('Errors');
$this->view->Errors = $flashMessenger->getMessages();
}
So for each controller I am able to use
$this->AddError($error);
And then I render $error in layout.
So I want not to deal with flashMesenger in every controller.
but I have to execute the activateErrors when each action is executed.
for example
I have an controller test
class TestController extends MyController {
public function indexAction() {
$this->AddError("Error 1");
$this->AddError("Error 2");
$this->activateErrors();
}
public function index1Action() {
$this->AddError("Esdsd 1");
$this->AddError("sddsd 2");
$this->activateErrors();
}
}
Is there a way that I could execute this activateErrors in each action for every controller at the end of action without duplicating the code.
I mean I do not want to include this code at every action. Maybe there is a way to include it in my abstract class MyController.
Anybody any Idea?
thanks
What about using a postDispatch hook, in your parent MyController ?
Quoting that page :
Zend_Controller_Action specifies two
methods that may be called to bookend
a requested action, preDispatch() and
postDispatch(). These can be useful in
a variety of ways: verifying
authentication and ACL's prior to
running an action (by calling
_forward() in preDispatch(), the action will be skipped), for instance,
or placing generated content in a
sitewide template (postDispatch()).
Maybe this might do the trick ?
I actually contributed an enhancement to FlashMessenger which provides a lot of the functionality you're looking for.