ASP NET MVC Razor pass model to layout

Question

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

User · Answer

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.

User · Answer

From above only  but simple implementation Index Page  model CMS Models IndexViewModel          ViewBag PageModel   Model         Layout Page        var Model    CMS Models IndexViewModel ViewBag PageModel

User · Answer

Let s assume your model is a collection of objects  or maybe a single object   For each object in the model do the following   1  Put the object you want to display in the ViewBag  For example     ViewBag YourObject   yourObject    2  Add a using statement at the top of  Layout cshtml that contains the class definition for your objects  For example    using YourApplication YourClasses   3  When you reference yourObject in  Layout cast it  You can apply the cast because of what you did in  2

User · Answer

A common solution is to make a base view model which contains the properties used in the layout file and then inherit from the base model to the models used on respective pages    The problem with this approach is that you now have locked yourself into the problem of a model can only inherit from one other class  and maybe your solution is such that you cannot use inheritance on the model you intended anyways   My solution also starts of with a base view model   public class LayoutModel       public LayoutModel string title                Title   title             public string Title   get       What I then use is a generic version of the LayoutModel which inherits from the LayoutModel  like this   public class LayoutModel lt T gt    LayoutModel       public LayoutModel T pageModel  string title    base title                PageModel   pageModel             public T PageModel   get        With this solution I have disconnected the need of having inheritance between the layout model and the model   So now I can go ahead and use the LayoutModel in Layout cshtml like this    model LayoutModel  lt  doctype html gt   lt html gt   lt head gt   lt title gt  Model Title lt  title gt   lt  head gt   lt body gt   RenderBody    lt  body gt   lt  html gt    And on a page you can use the generic LayoutModel like this    model LayoutModel lt Customer gt         var customer   Model PageModel      lt p gt Customer name   customer Name lt  p gt    From your controller you simply return a model of type LayoutModel   public ActionResult Page         return View new LayoutModel lt Customer gt  new Customer     Name    Test      Title

User · Answer

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

User · Answer

For example   model IList lt Model User gt          Layout    Views Shared SiteLayout cshtml       Read more about the new  model directive

User · Answer

There is another way to archive it    Just implement BaseController class for all controllers  In the BaseController class create a method that returns a Model class like for instance     public MenuPageModel GetTopMenu           var m   new MenuPageModel           populate your model here     return m          And in the Layout page you can call that method GetTopMenu       using GJob Controllers   lt header class  header-wrapper border-bottom border-secondary  gt     lt div class  sticky-header  id  appTopMenu  gt                var menuPageModel     BaseController this ViewContext Controller  GetTopMenu                 Html Partial   TopMainMenu   menuPageModel     lt  div gt   lt  header gt

User · Answer

Add a property to your controller  or base controller  called MainLayoutViewModel  or whatever  with whatever type you would like to use  In the constructor of your controller  or base controller   instantiate the type and set it to the property  Set it to the ViewData field  or ViewBag  In the Layout page  cast that property to your type    Example  Controller   public class MyController   Controller       public MainLayoutViewModel MainLayoutViewModel   get  set         public MyController                 this MainLayoutViewModel   new MainLayoutViewModel     has property PageTitle         this MainLayoutViewModel PageTitle    my title            this ViewData  MainLayoutViewModel     this MainLayoutViewModel             Example top of Layout Page     var viewModel    MainLayoutViewModel ViewBag MainLayoutViewModel      Now you can reference the variable  viewModel  in your layout page with full access to the typed object   I like this approach because it is the controller that controls the layout  while the individual page viewmodels remain layout agnostic   Notes for MVC Core  Mvc Core appears to blow away the contents of ViewData ViewBag upon calling each action the first time   What this means is that assigning ViewData in the constructor doesn t work   What does work  however  is using an IActionFilter and doing the exact same work in OnActionExecuting   Put MyActionFilter on your MyController   public class MyActionFilter  Attribute  IActionFilter               public void OnActionExecuted ActionExecutedContext context                               public void OnActionExecuting ActionExecutingContext context                        var myController  context Controller as MyController               if  myController   null                                myController Layout   new MainLayoutViewModel                                                        myController ViewBag MainLayoutViewModel  myController Layout

User · Answer

Maybe it isnt technically the proper way to handle it  but the simplest and most reasonable solution for me is to just make a class and instantiate it in the layout   It is a one time exception to the otherwise correct way of doing it   If this is done more than in the layout then you need to seriously rethink what your doing and maybe read a few more tutorials before progressing further in your project   public class MyLayoutModel       public User CurrentUser           get                  get the current user                        then in the view            Or get if from your DI container     var myLayoutModel   new MyLayoutModel        in  net core you can even skip that and use dependency injection    inject My Namespace IMyLayoutModel myLayoutModel   It is one of those areas that is kind of shady  But given the extremely over complicated alternatives I am seeing here  I think it is more than an ok exception to make in the name of practicality  Especially if you make sure to keep it simple and make sure any heavy logic  I would argue that there really shouldnt be any  but requirements differ  is in another class layer where it belongs   It is certainly better than polluting ALL of your controllers or models for the sake of basically just one view

User · Answer

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  lt  DOCTYPE html gt     lt html gt    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       lt div class  row  gt    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

User · Answer

public interface IContainsMyModel       ViewModel Model   get       public class ViewModel   IContainsMyModel       public string MyProperty   set  get        public ViewModel Model   get   return this         public class Composition   IContainsMyModel       public ViewModel ViewModel   get  set        Use IContainsMyModel in your layout   Solved  Interfaces rule

User · Answer

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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