How to use dynamic model to partial view in mvc?

Question

In one time I need to return IndexParty model. Next I need to return IndexPerson model to a view.

 public Tuple<IndexParty> GetTuple(){
     var tuple = new Tuple<IndexParty>(model);
     return tuple;
 }

My .cshtml class is here

@model Mymodel

<div class="hide partial-view" id="partyrelationship">
    @Html.Partial("../Party/Index", Mymodel)
</div>

Then I want to load as partial view.How can I do this?

score 0 · Answer 1 · answered Jan 07 '14 at 12:10

0

You just need a PartyController.cs (inheriting from Controller) with an Index action that returns this model.

PartyController:

public ActionResult Index()
{
    return PartialView(GetTuple());
}

answered Jan 07 '14 at 12:10

SethMW

score 0 · Answer 2 · answered Jan 07 '14 at 12:12

0

@model dynamic

<div class="hide partial-view" id="partyrelationship">
    @Html.Partial("../Party/Index", Model)
</div>

I would also move your partial to it's own controller action rather than using the index action.

answered Jan 07 '14 at 12:12

hutchonoid

When I try this it gave this error message 'System.Web.Mvc.HtmlHelper' has no applicable method named 'Partial' – user2901979 Jan 07 '14 at 12:26
I would take a look at this http://stackoverflow.com/questions/4047543/render-partial-view-with-dynamic-model-in-razor-view-engine-and-asp-net-mvc-3 – hutchonoid Jan 07 '14 at 12:31

2 Answers2