ASP.NET MVC Razor pass model to layout-第2页回答

What I see is a string Layout property. But how can I pass a model to layout explicitly?

标签： c# .net asp.net-mvc razor

10条回答

2楼-- · 2020-01-23 16:19

this is pretty basic stuff, all you need to do is to create a base view model and make sure ALL! and i mean ALL! of your views that will ever use that layout will receive views that use that base model!

public class SomeViewModel : ViewModelBase
{
    public bool ImNotEmpty = true;
}

public class EmptyViewModel : ViewModelBase
{
}

public abstract class ViewModelBase
{
}

in the _Layout.cshtml:

@model Models.ViewModelBase
<!DOCTYPE html>
  <html>
  and so on...

in the the Index (for example) method in the home controller:

    public ActionResult Index()
    {
        var model = new SomeViewModel()
        {
        };
        return View(model);
    }

the Index.cshtml:

@model Models.SomeViewModel

@{
  ViewBag.Title = "Title";
  Layout = "~/Views/Shared/_Layout.cshtml";
}

<div class="row">

i disagree that passing a model to the _layout is an error, some user info can be passed and the data can be populate in the controllers inheritance chain so only one implementation is needed.

obviously for more advanced purpose you should consider creating custom static contaxt using injection and include that model namespace in the _Layout.cshtml.

but for basic users this will do the trick

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3楼-- · 2020-01-23 16:20

Seems like you have modeled your viewmodels a bit wrong if you have this problem.

Personally I would never type a layout page. But if you want to do that you should have a base viewmodel that your other viewmodels inherits from and type your layout to the base viewmodel and you pages to the specific once.

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放荡不羁爱自由

4楼-- · 2020-01-23 16:24

Why dont you just add a new Partial View with i's own specific controller passing the required model to the partial view and finally Render the mentioned partial view on your Layout.cshtml using RenderPartial or RenderAction ?

I use this method for showing the logged in user's info like name , profile picture and etc.

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兄弟一词,经得起流年.

5楼-- · 2020-01-23 16:24

old question but just to mention the solution for MVC5 developers, you can use the Model property same as in view.

The Model property in both view and layout is assosiated with the same ViewDataDictionary object, so you don't have to do any extra work to pass your model to the layout page, and you don't have to declare @model MyModelName in the layout.

But notice that when you use @Model.XXX in the layout the intelliSense context menu will not appear because the Model here is a dynamic object just like ViewBag.

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ASP.NET MVC Razor pass model to layout

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