I have strongly typed list view
I have custom html helper which must get IEnumerable<object>
does it possible to pass my model(#model IEnumerable<MvcApplication2.Models.UserViewModel>) to my html helper?
If your helper is defined like this:
public static IHtmlString SomeHelper(this HtmlHelper<IEnumerable<object>> html)
then it is not possible to call it like this:
#model IEnumerable<MvcApplication2.Models.UserViewModel>
#Html.SomeHelper()
If on the other hand it is defined like this:
public static IHtmlString SomeHelper(this HtmlHelper html, IEnumerable<object> model)
you could call it from your view and pass the model:
#model IEnumerable<MvcApplication2.Models.UserViewModel>
#Html.SomeHelper(Model)
Related
I'm trying to do a very simple Html helper in Asp.net MVC 5:
#helper GetMyTextArea()
{
#Html.TextArea("myText")
}
Then i try to include it in my view: (the helper is in a file called MyHelp.cshtml, located in the App_Code Folder)
#MyHelp.GetMyTextArea()
If i render my view now, i get following exception:
System.NullReferenceException:
System.Web.WebPages.HelperPage.Html.get returned null.
Anyone know this issue? I think i can work around it with a partial view but this shouldn't be a problem with a html helper.
There are certain limitations apply when using #helper in App_Code folder, such like no direct access to standard HtmlHelper methods (you need to pass it manually by including it in your method, e.g. #helper GetMyTextArea(HtmlHelper Html)). You can add #functions which enables direct access to HtmlHelper methods:
#functions {
public static WebViewPage page = (PageContext.Page as WebViewPage);
public static HtmlHelper<object> html = page.Html;
}
Hence, MyHelp.cshtml content should be like this:
// adapted from /a/35269446/
#functions {
public static WebViewPage page = (PageContext.Page as WebViewPage);
public static HtmlHelper<object> html = page.Html;
}
#helper GetMyTextArea()
{
#Html.TextArea("myText")
}
Afterwards, you can call the helper method in view page as #MyHelp.GetMyTextArea() without passing HtmlHelper manually.
Similar issues:
Razor: Declarative HTML helpers
Using #Html inside shared #helper in App_Code
I used the following tutorial to help me build an RSS Reader in my ASP.NET MVC3 Razor application:
http://weblogs.asp.net/jalpeshpvadgama/archive/2011/08/17/creating-basic-rss-reader-in-asp-net-mvc-3.aspx
However, unlike the tutorial example, I want the RSS feed to be displayed on every page, and have therefore added it to my layout file, /Views/Shared/_Layout.cshtml
I currently only have 2 views on my site, and to get the RSS Reader to work on both views I've got the following code in my HomeController:
public class HomeController : Controller
{
//
// GET: /Index/
public ActionResult Index()
{
return View(CT.Models.RssReader.GetRssFeed());
}
public ActionResult About()
{
return View(CT.Models.RssReader.GetRssFeed());
}
}
From my WebForms experience, I would simply add the RSS Reader code in my master page code behind, and it would automatically work on every page.
Is there a Controller for layout pages which allows me to do the same?
How can I get this to work on every call of the layout page, without having to return anything?
EDIT: Following #Sebastian's advice, I've now added this code to a Partial View, removed CT.Models.RssReader.GetRssFeed() from return View() and included this in my layout file:
#Html.Partial("_MyPartialView")
The code in this partial view is:
<ul>
#foreach (var item in Model)
{
<li>
#item.Title
</li>
}
</ul>
However, I'm not getting a runtime error:
Object reference not set to an instance of an object.
It's erroring on the line #foreach (var item in Model)
You have to create a partial view and add functionality there.
Then in your layout, render this partial.
EDIT
Is your partial view really a partial view? The reason I said that is because you have "_" in front of the name which suggests that it might be a layout (might just be a naming convention).
To fix object reference error, you have to add the #Model declaration on top of your partial view.
Hope it helps.
UPDATE
In order to use different model in partial view, you need to explicitly declare which model you are going to use on render partialmethod.
#{Html.RenderPartial("../YourFeed", Model.YourFeedModel);}
Let me know if that resolved your issue.
The new error you are having is due to you not passing a Model to the partial view. You can do this with the second argument of the Html.Partial function...
Html.Partial("ViewName", MyModel);
As I think you are trying to do this in a Layout page you could also consider using a static reference to get your RSS feed. So forget about needing to pass in a Model and in your partial have:
#foreach (var item in RssRepository.GetFeed())
{
<li>
#item.Title
</li>
}
this like to a class something like...
public static RssRepository
{
public static MyModel GetFeed()
{
return new MyModel();//<- return what you would normally pass as a Model for RSS feeds
}
}
Hope that all makes sense
Ok.. here we go.. the weirdest and most confusing question of the month :)
I would like to create a HtmlHelper that some how renders html, but uses a partial view for its template of how the html should be rendered, so to put it more simple.. I would like to do exactly the same as a "normal" Controller and view does.. get some data, pass it to the view and then render the html, but in this case I would like to pass some data to a partial view, and then get the returned html as a string and then return that html from a HtmlHelper method...
In this way I would like to be able to write for instance #Html.HeadMenu, that then would return the html for the headmenu, but I would also be able to at anytime without recompiling be able to change the html.. since its all in a partial view.. and I wont have to worry about any server-side things.. and I will also get the benefit of the intellisense since my method will show up in #Html.
I hope you will understand this..since its kind of hard to explain..
Thanks in advance!
How about the Partial HTML-extension method, it sounds like what you are trying to achive right?
#{
var htmlString = Html.Partial("YourPartialViewName").ToString();
}
It also has an overload so that you can pass a model to the partial view:
#{
var htmlString = Html.Partial("YourPartialViewName", partialViewModel).ToString();
}
You could be looking for the Html.RenderAction(actionName, controllerName, routeValues) method.
I would do it this way
Define data that you imagine to pass to your htmlhelper
public class HeadMenuViewModel
{
public string SomeProperty {get;set;}
}
Define view named HeadMenuViewModel.cshtml in Views/Shared/DisplayTemplates
#model HeadMenuViewModel
<div>
////
</div>
From now, you can display your data using
#Html.DisplayFor(model => model.HeadMenu)
And you could write named shortcut-extension for it
using System.Web.Mvc.Html;
...
public static MvcHtmlString HeadMenu<TModel, TProperty>(this HtmlHelper<TModel> helper, Expression<Func<TModel, TProperty>> expression)
{
return helper.DisplayFor(expression);
}
Now, change your HeadMenuViewModel.cshtml everytime you need to
How can I create an individual controller and model for a partial view? I want to be able to place this partial view any where on the site so it needs it's own controller. I am current rendering the partial as so
#Html.Partial("_Testimonials")
Why not use Html.RenderAction()?
Then you could put the following into any controller (even creating a new controller for it):
[ChildActionOnly]
public ActionResult MyActionThatGeneratesAPartial(string parameter1)
{
var model = repository.GetThingByParameter(parameter1);
var partialViewModel = new PartialViewModel(model);
return PartialView(partialViewModel);
}
Then you could create a new partial view and have your PartialViewModel be what it inherits from.
For Razor, the code block in the view would look like this:
#{ Html.RenderAction("Index", "Home"); }
For the WebFormsViewEngine, it would look like this:
<% Html.RenderAction("Index", "Home"); %>
It does not need its own controller. You can use
#Html.Partial("../ControllerName/_Testimonials.cshtml")
This allows you to render the partial from any page. Just make sure the relative path is correct.
If it were me, I would simply create a new Controller with a Single Action and then use RenderAction in place of Partial:
// Assuming the controller is named NewController
#{Html.RenderAction("ActionName",
"New",
new { routeValueOne = "SomeValue" });
}
The most important thing is, the action created must return partial view, see below.
public ActionResult _YourPartialViewSection()
{
return PartialView();
}
You don't need a controller and when using .Net 5 (MVC 6) you can render the partial view async
#await Html.PartialAsync("_LoginPartial")
or
#{await Html.RenderPartialAsync("PartialName");}
or if you are using .net core 2.1 > you can just use:
<partial name="Shared/_ProductPartial.cshtml"
for="Product" />
Html.Action is a poorly designed technology.
Because in your page Controller you can't receive the results of computation in your Partial Controller. Data flow is only Page Controller => Partial Controller.
To be closer to WebForm UserControl (*.ascx) you need to:
Create a page Model and a Partial Model
Place your Partial Model as a property in your page Model
In page's View use Html.EditorFor(m => m.MyPartialModel)
Create an appropriate Partial View
Create a class very similar to that Child Action Controller described here in answers many times. But it will be just a class (inherited from Object rather than from Controller). Let's name it as MyControllerPartial. MyControllerPartial will know only about Partial Model.
Use your MyControllerPartial in your page controller. Pass model.MyPartialModel to MyControllerPartial
Take care about proper prefix in your MyControllerPartial. Fox example: ModelState.AddError("MyPartialModel." + "SomeFieldName", "Error")
In MyControllerPartial you can make validation and implement other logics related to this Partial Model
In this situation you can use it like:
public class MyController : Controller
{
....
public MyController()
{
MyChildController = new MyControllerPartial(this.ViewData);
}
[HttpPost]
public ActionResult Index(MyPageViewModel model)
{
...
int childResult = MyChildController.ProcessSomething(model.MyPartialModel);
...
}
}
P.S.
In step 3 you can use Html.Partial("PartialViewName", Model.MyPartialModel, <clone_ViewData_with_prefix_MyPartialModel>). For more details see ASP.NET MVC partial views: input name prefixes
I have problem with returning a partial view from a controller with different model than my main View. For example:
public ActionResult Index()
{
//myModel - get Some Types
return View(mymodel);
}
public PartialViewResult Categories()
{
//my another Model - get different Types
return PartialView(myanothermodel);
}
And then in Index View:
#Html.RenderPartial("Categories")
I get an exception saying that it is of the wrong type. It expects first type(mymodel) instead of second type.
Is it possible to return different types for view and its partial view?
Thanks for response.
It looks like you're trying to render the action, not the view.
Call #Html.Action("Categories").
When you are using Partial View only use
#Html.Partial("Categories", Model)
or a especific Model with your own data
#Html.Partial("Categories", Model.Category)
I just understood a bit how partial view works. In your and my case actually, no need to define the Categories() action if you think the logic to get myanothermodel can be done in Index() action.
So I have mymodel.myanothermodel assigned in the Index() action and then in the strongly typed Index.cshtml I used this: (assume myanothermodel is Categories)
#{Html.RenderPartial("Categories", Model.Categories);}
alternatively:
#Html.Partial("Categories", Model.Categories)
Note that always use .RenderPartial() rather than .Partial() for best performance in .cshtml view. I used Model.Categories instead of mymodel.Categories because of the strongly typed Index.cshtml already have #model mymodel in the begining of the file.
In my practise I have the models like:
Model.Departments - IList<DepartmentModel>
Model.SelectedDepartment - DepartmentModel
Model.Employees - IList<EmployeeModel>
which is used in:
#{Html.RenderPartial("DepartmentMenu", Model.Departments);}
<div id="employeeViewContainner">
#foreach (var emp in Model.Employees)
{
Html.RenderPartial("CardView" + Model.SelectedDepartments.Name, emp);
}
</div>
This will render employee list with different skin for different department.