CODEWITHSUNDEEP
asp.net-mvcuser-controlsstrong-typingviewdatapartials
Ayo
Ayo

Reputation: 1228

strongly-typed partial views MVC RC1

having a problem passing ViewData.Model to the partial views. It always is defaulting to null even if I equate it to a result query. I cannot access the strongly typed data because the Model is null. My current code is this,

ViewPage

<asp:Content ID="Content2" ContentPlaceHolderID="MainContent" runat="server">
    <% Html.RenderPartial("header", this.ViewData.Model); %>
    <% Html.RenderPartial("test", this.ViewData.Model); %>
    <div id="userControls">
    </div>
</asp:Content>

UserControl - header

<%@ Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<testMVCProject.Models.information>" %>
<h2>
    ACReport</h2>
<p>
    id:
    <%= Html.Encode(Model.id) %>
</p>
<p>
    type:
    <%= Html.Encode(Model.type) %>
</p>

UserControl - test

<%@ Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<testMVCProject.Models.information>" %>

        <%  using (Ajax.BeginForm(
            "pressureV2",
            "Home",
            new { id = ViewData.Model.id },
            new AjaxOptions
            {
                UpdateTargetId = "userControls",
                HttpMethod = "GET"

            },
            new { @id = "genInfoLinkForm" }))
            {%>
        <%= Html.SubmitButton("hey", "Lol") %>

    <%} %>

Controller

public ActionResult header(int id)
        {
            var headerResults = from c in db.information
                                where c.id == id
                                select new information
                                {
                                    id = c.id,
                                    type = c.type
                                };
            ViewData.Model = headerResults.FirstOrDefault();
            return View(ViewData.Model);
        }

public ActionResult pressureV2(int id)
        {
            var pressureVResults = from c in db.pressure_volume_tests
                                   where c.id == id
                                   select new pressureVT
                                   {
                                       bottomCVP = c.bottom_CVP,
                                       topCVP = c.top_CVP
                                   };

            ViewData.Model = pressureVResults.FirstOrDefault();
            return View(ViewData.Model);
        }

Upvotes: 0

Views: 2718

Answers (6)

Adam
Adam

Reputation: 1193

I found this worked for me, reference the partial as you do, like so.

...form
    @Html.Partial("_AboutYou", Model.AboutYou);
 ..end form

within the partial view at the top...

@model <namespace1>.<namespace2>.<namespace3>.CustomerInfo.AboutYou
    @{

        ViewData.TemplateInfo.HtmlFieldPrefix = "AboutYou";

        if (this.ViewContext.FormContext == null)
        {
            this.ViewContext.FormContext = new FormContext();
        }
    }

Upvotes: 0

alexl
alexl

Reputation: 6851

You have a misunderstanding of the use of Html.RenderPartial helper. When you use the RenderPartial you will show the view without requesting the model from the controller.

So you have to refactor your ViewPage and pass the good Model to your usercontrols:

Exemple:

Controller:

ActionResult MainView()
{
    var mainviewobj = new MainViewObject();

    var headerResults = from c in db.information
                                where c.id == id
                                select new information
                                {
                                    id = c.id,
                                    type = c.type
                                };

    mainviewobj.info = headerResults.FirstOrDefault();

    return view(mainviewobj);   
}

View Code:

<asp:Content ID="Content2" ContentPlaceHolderID="MainContent" runat="server">
    <% Html.RenderPartial("header", this.ViewData.Model.info); %>
    <% Html.RenderPartial("test", this.ViewData.Model.info); %>
    <div id="userControls">
    </div>
</asp:Content>

View Code Behind

public partial class MainView : ViewPage<MainViewObject>
{
}

Now the Model will not be null in your usercontrol. But remember the usercontrol rendering partially dun execute the code in the controller So you dun need the public ActionResult header(int id) in your Controller

Hope this helps.

Upvotes: 1

Justin
Justin

Reputation: 125

The Controller doesn't get called when you RenderPartial - it is bypassed and the view is rendered directly. So whatever you want to pass in as a model needs to be done from the calling View.

Upvotes: 0

liammclennan
liammclennan

Reputation: 5368

In the comments you have said that the view is not strongly typed. Because of that:

<% Html.RenderPartial("header", this.ViewData.Model); %>
<% Html.RenderPartial("test", this.ViewData.Model); %>

will not work. If you strongly type your view to testMVCProject.Models.information and then pass an instance of that type from your constructor it will work.

Controller:

public ActionResult ShowAView()
{
    Return View("WhateverYourViewIsCalled", new information());
}

Upvotes: 2

eulerfx
eulerfx

Reputation: 37749

Have you tried making the ViewPage generic as well?

Upvotes: 0

antonioh
antonioh

Reputation: 2944

I believe the problem might be that you're missing an element in the form with the name "id" so the parameter of the Action method is never populated with a value?

That way the query would always return null with the FirstOrDefault, hence the null Model.

Just my guess...

Upvotes: -1

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