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Render a View inside a View in Asp.Net mvc

How do I render a full fledged view (not partial view) inside another view?
Scenario, I have different controller and want the exactly same view to render which is already there under other controller with different layout.
I have Wishlist page in Home Controller which shows list of added products, and when user logged in , when I click on wish list it also show me navigation when user is signed in.
How would I do that??

Not many developers know about this but you can use RenderPage, it's specifically designed for that purpose(to render an MVC view inside another view)
#RenderPage("~/Views/Shared/SampleView.cshtml")

You can still create or use a partial view with its own controller and use the RenderAction()
[ChildActionOnly]
public ActionResult ActionPartialView(string p1)
{
//code...
return PartialView();
}
The above code can be in any controller, its own controller, just call it in razor with that controller.
Razor:
#{ Html.RenderAction("Index", "Home"); }
Hope that helps

In MVC architecture, can a controller and model manage more than one View?

Can I create 2 views for a model and controller?
My current application is MVC complaint , and it has a single view.
I need to create a second GUI, that fetches few information from the model (updated from the first GUI data) , update it and display back in the first GUI.

You can have a different view per action. Based on your description, this seems to be what you are looking for. You have different actions for the same model, it's ok.
You can add a new action for the new information that you want to update and click on the right button of the mouse and click "add view". It will add a new view for that action.

The most common way to do this is to create other actions, one for each View you want to display.
You can have as many Views as you like, as long as you can route between them.
As default, the line return View(); or return View(model) will look for a view in this path: /Views/{ControllerName}/{ActionName}.
You can also specify the view name, as Controller.View() also accepts a string as the view name.
Knowing this, you could display different views from a single action, according to parameters passed to your action.
Example:
public ActionResult Example(bool a)
{
if (a) return View("a");
else return View("b");
}
this will call the view /Views/{ControllerName}/a.cshtml if a is true and /Views/{ControllerName}/b.cshtml if a is false.
You can also call other partial views or actions from your first view, using html helpers: #Html.Partial({ViewName}), #{Html.RenderPartial({ViewName});} or #{Html.RenderAction({ActionName});}
EDIT:
View() also searchs for /Views/Shared/{ActionName}

MVC Partial view to Display list

I have an Action result that returns a list
public ActionResult GetData(Profiles profiles)
{
Vertical Vdata = new Vertical();
List<Ver> Vertical = new List<Ver>();
//Code to fill list
return View(Vertical);
}
And then a partial view to display the List
#model IEnumerable< List<VerticalContainer.Models.Vertical>>
#foreach(var item in Model)
{
<span>#item.name</span>
}
I'm not sure how to render the partial view from my main view
#Html.Action("GetData")??
What do I pass with the Html.Action? or Should I use Partial/RenderPartial?

1- the correct way to render actions is as per the following:
#Html.RenderAction("ACTION_NAME","CONTROLLER_NAME")
where you replace both action name and controller name with the correct values according to your solution.
2- For the passed model, if this action is being rendered inside a view that has a Model property of Profiles
, then you don't have to specify the model to be passed to the action, as it will implicitly read it from the parent view.
if this is not the case, then you will need to store your Profiles values inside a medium variable (for example inside a ViewBag property) and then you will pass it when calling the action, so it should work on this passed model istead of working on the parent view one.
example: #Html.RenderAction("ACTION_NAME","CONTROLLER_NAME", YOUR_MODEL_VALUE_HERE)
Suggestion: if this parital view will just render the list passed, you can skip creating the mentioned action, and just make a call which can render the partial view, passing to it the correct list so it can work with.
To make it more clear, using your question I can see that the action named GetData is just constructing the list called Vertical from the Profiles model passed, so you can construct this list in your original action (Index, Details, WHAT_EVER_THE_NAME_IS) and store it in a ViewBag, then you can call RenderParial instead of RenderAction and this should result in the same output as your mentioned scenario.
example: #{ Html.Partial("ViewName", YOUR_MODEL_HERE);}

#{Html.RenderAction("actionName","controllerName");}

From your main view, you may simply render your Partial view as:
#Html.Partial("_PartialViewName", VerticalList);
VerticalList in this case will be the list of type VerticalContainer.Models.Vertical, that you populated in controller action.

Re-using a partial view (ascx) across different Controllers and Views

Let's assume I have 2 Controllers, TopicsController and PostsController.
For each controller, I have a couple of views (Index & Details).
The Topic (Index) view inherits System.Web.Mvc.ViewPage<IEnumerable<MessageBoard.Models.Topic>>
The Topic (Details) view inherits System.Web.Mvc.ViewPage<MessageBoard.Models.TopicFormViewModel>
I'm using a TopicFormViewModel because I'm sending additional data along with the Model.
The Post (Details) view just inherits System.Web.Mvc.ViewPage<MessageBoard.Models.Post>
Now, I've created a partial view (CreatePost.ascx) which is (obviously :p) used to create a new Post. I want to be able to re-use this control on all of the views you see above.
Update
I've tried rendering the partial view using <% Html.RenderPartial("New"); %> from my Topics/Index.aspx View, but that results in an exception
The model item passed into the dictionary is of type 'System.Data.Linq.Table`1[MessageBoard.Models.Topic]', but this dictionary requires a model item of type 'MessageBoard.Models.Post'.
Now the problem is that my partial view (CreatePost.ascx) accepts a System.Web.Mvc.ViewUserControl<MessageBoard.Models.Post> and I'm not sure how to pass that from all my views above.
I'm also unsure know how to submit the .ascx values to a certain URL (i.e. /Topics/1/CreatePost), how do I tell the submit button to post to that URL?
Thanks in advance,
Marko

Ciao Marko,
Now the problem is that my partial
view (CreatePost.ascx) accepts a
System.Web.Mvc.ViewUserControl
and I'm not sure how to pass that from
all my views above.
I am not sure I understand what do you mean by "how to pass that from all my views above" but I am sure that you dont have to pass an instance of Post from your views. What is going on is that from your views you will invoke a controller action that creates the Post model object and then bind it to the CreatePost.ascx partial.
I'm also unsure know how to submit the
.ascx values to a certain URL (i.e.
/Topics/1/CreatePost), how do I tell
the submit button to post to that URL?
You have two options:
Inside your CreatePost.ascx partial you are probably using a form.
<% using (Html.BeginForm("action", "controller", FormMethod.Post, new {} )) { %>
If you use in the way I am showing you can change the first and the second params respectively to the names of the Action and the Controller that would habndle your submit.
The second option is using jQuery. Simply set an ID for your form and then
$("#myForm").submit(function(event) {
//post an ajax request to the server
});
Hope this helps!
P.S. To be able to reuse your CreatePost.ascx partial place it inside the shared view folder (where your master page is).

In regards to reusing a partial view which is not in the same view folder, use the following and pass in the model required, alternatively you can define a custom route for it.
<% html.RenderPartial("~/Views/<ControllerName>/<PartialViewName>.ascx", <model>);

#Marko
Another way would be to have in the PostController an Action like the following:
[HttpGet]
public ActionResult CreatePost( int topicId ) {
PostModel pm = _manager.CreateDefaultPost();
pm.TopicID = id;
return PartialView( "CreatePost", pm );
}
Then wherever you want to create a Post you can simply call this action that returns the strong-typed view for your new post.
Even if you have a supplementary http call over the network IMO this solution has the advantage to keep the initialization code of a new Post centralized to a single place.
From the View "point-of-view" the call to the action can be done when the user press the "New Post" button and then inject the received markup inside a modal dialog or in a place of your pleasure in the current page.
Hope it helps!

Do I need to have a View Page for every Action in ASP.NET MVC?

I have a customer index page that displays customer lists. I have a search functionality within this page and I want the url to be http://mysite/search?id=434 when i perform the search. The index page will also display the search result.
Thanks

public class CustomerController : Controller
...
public ActionResult Search(int id)
{
ViewData["SearchResult"] = MySearchBLL.GetSearchResults(id);
return View("Index");
}
...
Hope this helps

No, you shouldn't have. Just use View("ViewName"); in the controller to show the appropriate view in other actions.

In the URL that you suggest, you should have a Controller method called 'search' and using the 'index' view for that controller.
If thats the case, you can post it back to the same action and in the Controller have different sets of code for the 'GET' and 'POST' to give the functionality that you are looking for.

Have the HTML form post to the same action that rendered it. That action can decide if it's a non-search (first visit) or a search results rendering. That action can populate ViewData with the appropriate data. That is, if you want to do that.
You can also have two views, really easily. And the action can transparently decide which one to render.