CODEWITHSUNDEEP
asp.net-mvc-3partial-views
David Abraham
David Abraham

Reputation: 197

Unable to pass model into partial view inside layout pages

I have a layout page within that layout I have used a partial view, the partial view contains a menu feature that I have built, I split the menu into a partial view to ensure it is easy to maintain. The menu has been purposely placed in the layout because it is used across every page, however there are conditional elements in the menu whereby some options only appear on certain pages.

I have stumbled upon what will be an issue for me moving forward, the menu uses ajax calls to render partial views containing the content (reducing page loads) I was just working on a page which contains a company, the company contains a list of contacts, the menu option when clicked should display the list of contacts. I have already loaded the contact list under the company model, but! I can’t access it from my new partial view that should render the contact list because the menu is a partial view that is contained within the layout page and as such cannot accept a model, so I cannot pass a model into the partial view I am trying to load because the menu partial view sits in the layout page.

This is a sticky situation, I obviously could change the layout to render a new section to contain the menu so I can pass a new view model into it but then every single page I build needs to reference the menu (what a pest!) I must be missing something here (considering this is my first MVC3 application that is likely). Any suggestions?

Edit: I took this further on my own, in short my layout page will always be able to access the model of the page that consumes it, as such my partial view which contains the menu can also access that data. I wrote some conditional logic in my menu partial view that checks the page and then passes in data as required.

<div class="menu">    
    <ul>
        <li><a href="@Url.Action("Create", "Contact")">New Contact </li>
        <li><a href="@Url.Action("Index", "Contact")">Contact List </a></li>
    </ul>
    @if (Request.Url.PathAndQuery.Contains("/Contact/Details/"))
    {
        <ul>
            <li>@Html.ActionLink("New Activity", "Create", "Activity", new { companyid = 0, contactid = Model.contact.id }, null)</li>
        </ul>
    }
</div>

The above is a small sample of the menu partial view but contains one example where the menu is built for the contact/details page and is able to pass in the model.contact.id. It works in that my menu and my layout do not explicitly contain a model, but it does not feel very tidy.

Upvotes: 2

Views: 1836

Answers (1)

DMulligan
DMulligan

Reputation: 9073

If I'm understanding your question correctly, your issue is that you don't think your Partial View can have a model because you don't want your layout to have a model. So the question is how can you get a model into your layout without needing every single action to extend this base model type your layout would use.

1) Instead of using Html.Partial in your layout for the menu use Html.Action where you'll then have an action method that fetches the menu data.

2) Write a custom WebViewPage and include a property that has something like

return ((BaseController)ViewContext.Controller).MenuData;

now you don't even need a model in your partial view, it can access the data directly.

Both of these require having a Menu property containing all menu information available in your base model, but if every page in your websites going to need to access this data, then that seems appropriate.

Edit: In response to your tidiness concern, it sounds like you want sections, which give you the ability to customize pieces of your menu either in their appropriate view page or sub layout.

See http://weblogs.asp.net/scottgu/archive/2010/12/30/asp-net-mvc-3-layouts-and-sections-with-razor.aspx for an overview of sections and http://blogs.msdn.com/b/marcinon/archive/2010/12/15/razor-nested-layouts-and-redefined-sections.aspx for information of nested layouts/sections.

Upvotes: 5

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