I am using an edit template for Kendo scheduler called ScheduleEditorTemplate.cshtml which his located in Views/Shared/EditorTemplates
I have a partial view named _POC.cshtml and I need to display that partial view inside the ScheduleEditorTemplate.cshtml. This is a read-only partial view, the user will only view what is there.
When I use the RenderPartial method as follows I get an "invalid template" exception when I try to open the ScheduleEditorTemplate.cshtml from the scheduler. It works without the partial view included.
How can I render a partial view with a different view model into an editor template?
Here's my code to render the partial view within the ScheduleEditorTemplate.cshtml
#{Html.RenderPartial("_POC");}
I am already using this partial view in another view so I know it loads and works correctly there.
I tried this also with the same exception generated:
#Html.Partial("~/Views/EmployeeSchedule/_POC.cshtml")
I can share the code from _POC.cshtml but there's nothing special there; just kendo controls.
#{Html.RenderPartial("_POC");}
And use Ajax to load this partial view:
<script type="text/javascript">
$(document).ready(function () {
$("#div").load('#(Url.Action("POC","ControllerName"))', function(){});
});
</script>
Related
In the shop's customer section I would like to render the user account menu (which we can see by default when reaching /my-account URL) as a side column on some others controllers like the ones associated to "/my-adresses", "/identity" pages ..
I thought I would have to create another controller which purpose would be to gather menu infos and only render the menu <ul> list. Then I could override Controllers such as MyAccountController, IdentityController to include this former Ctrl and then render its content as part of the views of those two other controllers views.
So how one can load a specific controller from another in order to render shared views between pages ? Which is the right/clean way to do that ?
I heard about $this->getController() but I did not find any snippet or implementation of what I'd like to achieve. I new to Prestashop but even if the code seems clear, I don't get the point here.
Thank you !
After getting a bit deeper into sources, I ended up moving the menu (initially part of the MyAccountControllerCore alias my-account template) into a brand new Controller "MyAccountMenuController", in /override/controllers/front/.
<?php
// In /override/controllers/front/MyAccountMenuController.php
// The "exposer" controller
public function display()
{
// Do what ever you want to pass specific variables
if (! $this->template) {
throw new Exception(get_class($this) . '::display() : missing template.');
}
$this->context->smarty->display($this->template);
return true;
}
I am now able to import this menu by adding the following snippet inside the initContent() method of each controller that are part of the customer account section (this means that each of them must be overridden, for more info about overriding controllers, see documentation):
<?php
// In /override/controllers/front/MyAccountController.php
// The "consumer" controller
public function initContent() {
// ...
// Importing the customer area's menu
$menuController = $this->getController('MyAccountMenuController');
ob_start();
$menuController->run();
$this->context->smarty->assign('myAccountMenu', ob_get_clean());
}
I don't think the original purpose of getController method (located in ControllerCore class) was meant to include another controller output, at least in Prestashop 1.5. Still, to me this approach is far cleaner than duplicate views code.
If you have a better (cleaner) approach to implement such a mechanism, please let me know !
Any thoughts ?
How do I post from one controller into another view?
I have a Review model and a Product model. The Review form is displayed in the Product view through a widget, but how do I submit the form itself? Right now, it doesn't do anything. I can submit through review/create, but not through the Product View.
Or am i suppose to do the post in the widget?
You can achieve it if you put code like below on components/ReviewWidget.php . I supposed you have Review as model and its respective controller and views file on default locations.
<?php
class ReviewWidget extends CWidget{
public function init() {
return parent::init();
}
public function run(){
$model = new Review;
if (isset($_POST['Review'])) {
$model->attributes = $_POST['Review'];
$model->save();
}
$this->renderFile(Yii::getPathOfAlias('application.views.review'). '/_form.php',array(
'model' => $model,
));
}
}
Then, call above widget on any where on view like below ,
<?php $this->widget('ReviewWidget'); ?>
It will handle item creation only. You have to create code to item update by yourself.
In your controller action you must use function renderPartial
$this->renderPartial('//views/reviw/_form',array('data' => $data ) );
First argument of this function is used to determine which view to use:
absolute view within a module: the view name starts with a single slash '/'. In this case, the view will be searched for under the
currently active module's view path. If there is no active module,
the view will be searched for under the application's view path.
absolute view within the application: the view name starts with double slashes '//'. In this case, the view will be searched for
under the application's view path. This syntax has been available
since version 1.1.3.
aliased view: the view name contains dots and refers to a path alias. The view file is determined by calling
YiiBase::getPathOfAlias(). Note that aliased views cannot be themed
because they can refer to a view file located at arbitrary places.
relative view: otherwise. Relative views will be searched for under the currently active controller's view path.
Also you can use this function in your views. But the most convenient way to reuse views is to create widgets.
I have some issues with rendering strongly type partial view in layout view.
partial view (menu) is band with model Menu
this partial view I want to render in _layout.chtml, so it 'll be avliable in all views.
I want to make avalible the partial view(menu) in all pages/view. the problem I face is where to put the action for partial view to populate it from DB on page load.
thanks
---------------------- My code is--------------------
partial view inside shared folder.
#model List<Menu>
#foreach(var item in Model){// here is the html/model item inside to display}
--------------------------------------
HomeView.chtml inside home folder
#model List<homemodel>
.... here goes html code/ plus homemodel loop/data etc.
------------------------------
HomeController{
public ActionResult HomeView()
{
.........return view();
}
public PartialViewResult partialmenu()
{
// data from db
return partialview(partialobject as list);
}
------------------------
layoutview.chtml
--html code---
{# Html.renderpartial("partialview");}
.. html code...
I want to make available the partial view(menu) in all pages/view. the problem I face is where to put the action for partial view to populate it from DB on page load.
Write in layout: Html.RenderAction('MyMenu') OR Html.Action('MyMenu') and then populate it from any source. Your action will return strongly typed model.
where to put the action for partial view to populate it from DB on
page load.
every view has it's own controller..regardless if it is partial or not...therefore you could populate your view on it's own controller...
is there a way to change stylesheet file dynamically depending on if the document is in edit or read mode?
What I would like to do is to add the following code to the "compute value" option of the resource href property:
if(document.isEditable()){
return "style_edit.css"
}
else{
return "style_read.css"
}
My main problem with this is that when the page loads, it gives the error "document not found". This is probably because when the page loads, there is only a view that includes the documents and when the user clicks a document id, then the custom control with the binded document appears. I don't know how to make the binded to custom control document available on load of the page.
Edited:
I tried a try/catch block and now the xpage opens without displaying an error. But although the custom control is refreshed, the css file does not change, althoug I use compute dynamically and not compute on load
Thank you in advance!
You can set the resource href attribute as computed. For this go to All Properties of XPage "basics > resources > styleSheet". Here you can compute the href attribute with your JavaScript code. So your resource in XPage source would look something like this
<xp:this.resources>
<xp:styleSheet>
<xp:this.href><![CDATA[#{javascript:if (document.isEditable()) {
return "style_edit.css";
} else {
return "style_read.css";
}}]]></xp:this.href>
</xp:styleSheet>
</xp:this.resources>
To access the data source from custom control you can use the global variable currentDocument instead of document.
Why force user to download separate files on edition when You can simply add computed styleClass to some panel/component:
<xp:panel>
<xp:this.styleClass><![CDATA[#{javascript:return document.isEditable()?"docEditMode":"docReadMode";}]]></xp:this.styleClass>
</xp:panel>
and use it as a selector inside style.css
In Dojo, I am trying to extend dijit.Dialog using templates. When I instantiate it, I get only the text in the dialog box, without the borders or close button. Is there some additional step I need to do to get it fully initialized?
My template is in template.html, it looks like so:
<div dojoType="dijit.Dialog" id="dynFilter" jsId="dynFilter">
"Dynamic Dialog"
</div>
Here is the dojo.declare:
dojo.declare(
"template.dialog", // class name
[dijit._Widget, dijit._Templated, dijit.Dialog], // parent classes
{
templateString : dojo.cache("autonomics", "template.html"),
}
);
After I instantiate it, I call .startup(), which doesn't seem to do anything, then .show(), which does place it on the page, missing most of its functionality.
var dialog = new template.dialog();
dialog.startup();
dialog.show();
What am I missing?
You overwrite the template of the original dijit/Dialog when subclassing.
Have a look to my answer to Dojo Dialog with confirmation button which solves the issue you are experiencing. Or go directly to working example at jsFiddle: http://jsfiddle.net/phusick/wkydY/