@model for _layout.cshtml on MVC4?

Question

I was wondering if there\'s a way to specify a model for the _layout.cshtml file, i\'ve seen lots of posts with the basic same question with people replying with \"alternati

Answer 1

You should delegate the parts of your layout that "need a model" to a separate controller using partial views and RenderAction:

@Html.RenderAction("SomeAction", "LayoutController")

Have LayoutController.SomeAction return a PartialViewResult, which you can then strongly type to a model.

Answer 2

Even though you already accepted an answer, based on your saying you are just pulling an image URL you should do it using JQuery, not a model.

This code is untested, apologies for that. Feel free to point out if I typed a bug. The HTML element containing the background image has the id="url" attribute so the selectors work.

Controller

[HttpGet]
public string GetSessionUrl()
{
    //logic to detmine url
    return url;
}

JQuery

$(document).ready(function () {
    var $url = $('#url');
    var options = {
        url: "/Home/GetSessionUrl",
        type: "get",
        async:false
    };

    $.ajax(options).done(function (data) {
        $url.attr('src', data);
    });
});

Answer 3

You can add BaseModel to _Layout.

@model BaseModel

Then all models inherit from that BaseModel class.

public class MyModel : BaseModel
{
}

As others stated, it is not a good practice. If your model forgets to inherit from BaseModel, it'll throws exception at run time. However, it is up to you.

Answer 4

In BaseController you can declare any model as property.

public class BaseController : Controller
{
    public BaseController ()
    {
        MyTag = new TagModel ();  // or get db, take any value from there           
    }

    public TagModel MyTag { get; set; }
}

In action:

ViewBag.MyTag = MyTag ;

And in _Layout.cshtml, you can use

@{
  var myTag = (TagModel)ViewBag.MyTag;
}