[asp.net-mvc-4] Pass data to layout that are common to all pages

I have a website which have a layout page. However this layout page have data which all pages model must provide such page title, page name and the location where we actually are for an HTML helper I did which perform some action. Also each page have their own view models properties.

How can I do this? It seems that its a bad idea to type a layout but how do I pass theses infos?

Why hasn't anyone suggested extension methods on ViewData?

Option #1

Seems to me by far the least intrusive and simplest solution to the problem. No hardcoded strings. No imposed restrictions. No magic coding. No complex code.

public static class ViewDataExtensions
{
    private const string TitleData = "Title";
    public static void SetTitle<T>(this ViewDataDictionary<T> viewData, string value) => viewData[TitleData] = value;
    public static string GetTitle<T>(this ViewDataDictionary<T> viewData) => (string)viewData[TitleData] ?? "";
}

Set data in the page

ViewData.SetTitle("abc");

Option #2

Another option, making the field declaration easier.

public static class ViewDataExtensions
{
    public static ViewDataField<string, V> Title<V>(this ViewDataDictionary<V> viewData) => new ViewDataField<string, V>(viewData, "Title", "");
}

public class ViewDataField<T,V>
{
    private readonly ViewDataDictionary<V> _viewData;
    private readonly string _field;
    private readonly T _defaultValue;

    public ViewDataField(ViewDataDictionary<V> viewData, string field, T defaultValue)
    {
        _viewData = viewData;
        _field = field;
        _defaultValue = defaultValue;
    }

    public T Value {
        get => (T)(_viewData[_field] ?? _defaultValue);
        set => _viewData[_field] = value;
    }
}

Set data in the page. Declaration is easier than first option, but usage syntax is slightly longer.

ViewData.Title().Value = "abc";

Option #3

Then can combine that with returning a single object containing all layout-related fields with their default values.

public static class ViewDataExtensions
{
    private const string LayoutField = "Layout";
    public static LayoutData Layout<T>(this ViewDataDictionary<T> viewData) => 
        (LayoutData)(viewData[LayoutField] ?? (viewData[LayoutField] = new LayoutData()));
}

public class LayoutData
{
    public string Title { get; set; } = "";
}

Set data in the page

var layout = ViewData.Layout();
layout.Title = "abc";

This third option has several benefits and I think is the best option in most cases:

• Simplest declaration of fields and default values.

• Simplest usage syntax when setting multiple fields.

• Allows setting various kinds of data in the ViewData (eg. Layout, Header, Navigation).

• Allows additional code and logic within LayoutData class.

P.S. Don't forget to add the namespace of ViewDataExtensions in _ViewImports.cshtml

You don't have to mess with actions or change the model, just use a base controller and cast the existing controller from the layout viewcontext.

Create a base controller with the desired common data (title/page/location etc) and action initialization...

public abstract class _BaseController:Controller {
    public Int32 MyCommonValue { get; private set; }

    protected override void OnActionExecuting(ActionExecutingContext filterContext) {

        MyCommonValue = 12345;

        base.OnActionExecuting(filterContext);
    }
}

Make sure every controller uses the base controller...

public class UserController:_BaseController {...

Cast the existing base controller from the view context in your _Layout.cshml page...

@{
    var myController = (_BaseController)ViewContext.Controller;
}

Now you can refer to values in your base controller from your layout page.

@myController.MyCommonValue

UPDATE

You could also create a page extension that would allow you to use this.

//Allows typed "this.Controller()." in cshtml files
public static class MyPageExtensions {
    public static _BaseController Controller(this WebViewPage page) => Controller<_BaseController>(page);
    public static T Controller<T>(this WebViewPage page) where T : _BaseController => (T)page.ViewContext.Controller;
}

Then you only have to remember to use this.Controller() when you want the controller.

@{
    var myController = this.Controller(); //_BaseController
}

or specific controller that inherits from _BaseController...

@{
    var myController = this.Controller<MyControllerType>();
}

what i did is very simple and it's works

Declare Static property in any controller or you can make a data-class with static values if you want like this:

public static username = "Admin";
public static UserType = "Administrator";

These values can be updated by the controllers based on operations. later you can use them in your _Layout

In _layout.cshtml

@project_name.Controllers.HomeController.username
@project_name.Controllers.HomeController.UserType

It's incredible that nobody has said this over here. Passing a viewmodel through a base controller is a mess. We are using user claims to pass info to the layout page (for showing user data on the navbar for example). There is one more advantage. The data is stored via cookies, so there is no need to retrieve the data in each request via partials. Just do some googling "asp net identity claims".

There's another way to handle this. Maybe not the cleanest way from an architectural point of view, but it avoids a lot of pain involved with the other answers. Simply inject a service in the Razor layout and then call a method that gets the necessary data:

@inject IService myService

Then later in the layout view:

@if (await myService.GetBoolValue()) {
   // Good to go...
}

Again, not clean in terms of architecture (obviously the service shouldn't be injected directly in the view), but it gets the job done.

if you want to pass an entire model go like so in the layout:

@model ViewAsModelBase
<!DOCTYPE html>
<html>
<head>
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta charset="utf-8"/>
    <link href="/img/phytech_icon.ico" rel="shortcut icon" type="image/x-icon" />
    <title>@ViewBag.Title</title>
    @RenderSection("styles", required: false)    
    <script type="text/javascript" src="http://code.jquery.com/jquery-1.8.3.min.js"></script>
    @RenderSection("scripts", required: false)
    @RenderSection("head", required: false)
</head>
<body>
    @Html.Action("_Header","Controller", new {model = Model})
    <section id="content">
        @RenderBody()
    </section>      
    @RenderSection("footer", required: false)
</body>
</html>

and add this in the controller:

public ActionResult _Header(ViewAsModelBase model)

You can use like this:

 @{ 
    ApplicationDbContext db = new ApplicationDbContext();
    IEnumerable<YourModel> bd_recent = db.YourModel.Where(m => m.Pin == true).OrderByDescending(m=>m.ID).Select(m => m);
}
<div class="col-md-12">
    <div class="panel panel-default">
        <div class="panel-body">
            <div class="baner1">
                <h3 class="bb-hred">Recent Posts</h3>
                @foreach(var item in bd_recent)
                {
                    <a href="/BaiDangs/BaiDangChiTiet/@item.ID">@item.Name</a>
                }
            </div>
        </div>
    </div>
</div>

Creating a base view which represents the Layout view model is a terrible approach. Imagine that you want to have a model which represents the navigation defined in the layout. Would you do CustomersViewModel : LayoutNavigationViewModel? Why? Why should you pass the navigation model data through every single view model that you have in the solution?

The Layout view model should be dedicated, on its own and should not force the rest of the view models to depend on it.

Instead, you can do this, in your _Layout.cshtml file:

@{ var model = DependencyResolver.Current.GetService<MyNamespace.LayoutViewModel>(); }

Most importantly, we don't need to new LayoutViewModel() and we will get all the dependencies that LayoutViewModel has, resolved for us.

e.g.

public class LayoutViewModel
{
    private readonly DataContext dataContext;
    private readonly ApplicationUserManager userManager;

    public LayoutViewModel(DataContext dataContext, ApplicationUserManager userManager)
    {
    }
}

Other answers have covered pretty much everything about how we can pass model to our layout page. But I have found a way using which you can pass variables to your layout page dynamically without using any model or partial view in your layout. Let us say you have this model -

public class SubLocationsViewModel
{
    public string city { get; set; }
    public string state { get; set; }
}

And you want to get city and state dynamically. For e.g

in your index.cshtml you can put these two variables in ViewBag

@model  MyProject.Models.ViewModel.SubLocationsViewModel
@{
    ViewBag.City = Model.city;
    ViewBag.State = Model.state;
}

And then in your layout.cshtml you can access those viewbag variables

<div class="text-wrap">
    <div class="heading">@ViewBag.City @ViewBag.State</div>
</div>

You can also make use of RenderSection , it helps to you to inject your Model data into the _Layout view.

You can inject View Model Data, Json, Script , CSS, HTML etc

In this example I am injecting Json from my Index View to Layout View.

Index.chtml

@section commonLayoutData{

    <script>

        var products = @Html.Raw(Json.Encode(Model.ToList()));

    </script>

    }

_Layout.cshtml

@RenderSection("commonLayoutData", false)

This eliminates the need of creating a separate Base View Model.

Hope helps someone.

I used RenderAction html helper for razor in layout.

@{
   Html.RenderAction("Action", "Controller");
 }

I needed it for simple string. So my action returns string and writes it down easy in view. But if you need complex data you can return PartialViewResult and model.

 public PartialViewResult Action()
    {
        var model = someList;
        return PartialView("~/Views/Shared/_maPartialView.cshtml", model);
    }

You just need to put your model begining of the partial view '_maPartialView.cshtml' that you created

@model List<WhatEverYourObjeIs>

Then you can use data in the model in that partial view with html.

I do not think any of these answers are flexible enough for a large enterprise level application. I'm not a fan of overusing the ViewBag, but in this case, for flexibility, I'd make an exception. Here's what I'd do...

You should have a base controller on all of your controllers. Add your Layout data OnActionExecuting in your base controller (or OnActionExecuted if you want to defer that)...

public class BaseController : Controller
{
    protected override void OnActionExecuting(ActionExecutingContext     
        filterContext)
    {
        ViewBag.LayoutViewModel = MyLayoutViewModel;
    }
}

public class HomeController : BaseController
{
    public ActionResult Index()
    {
        return View(homeModel);
    }
}

Then in your _Layout.cshtml pull your ViewModel from the ViewBag...

@{
  LayoutViewModel model = (LayoutViewModel)ViewBag.LayoutViewModel;
}

<h1>@model.Title</h1>

Or...

<h1>@ViewBag.LayoutViewModel.Title</h1>

Doing this doesn't interfere with the coding for your page's controllers or view models.

instead of going through this you can always use another approach which is also fast

create a new partial view in the Shared Directory and call your partial view in your layout as

@Html.Partial("MyPartialView")

in your partial view you can call your db and perform what ever you want to do

@{
    IEnumerable<HOXAT.Models.CourseCategory> categories = new HOXAT.Models.HOXATEntities().CourseCategories;
}

<div>
//do what ever here
</div>

assuming you have added your Entity Framework Database

You could create a razor file in the App_Code folder and then access it from your view pages.

Project>Repository/IdentityRepository.cs

namespace Infrastructure.Repository
{
    public class IdentityRepository : IIdentityRepository
    {
        private readonly ISystemSettings _systemSettings;
        private readonly ISessionDataManager _sessionDataManager;

        public IdentityRepository(
            ISystemSettings systemSettings
            )
        {
            _systemSettings = systemSettings;
        }

        public string GetCurrentUserName()
        {
            return HttpContext.Current.User.Identity.Name;
        }
    }
}

Project>App_Code/IdentityRepositoryViewFunctions.cshtml:

@using System.Web.Mvc
@using Infrastructure.Repository
@functions
{
    public static IIdentityRepository IdentityRepositoryInstance
    {
        get { return DependencyResolver.Current.GetService<IIdentityRepository>(); }
    }

    public static string GetCurrentUserName
    {
        get
        {
            var identityRepo = IdentityRepositoryInstance;
            if (identityRepo != null)
            {
                return identityRepo.GetCurrentUserName();
            }
            return null;
        }
    }
}

Project>Views/Shared/_Layout.cshtml (or any other .cshtml file)

<div>
    @IdentityRepositoryViewFunctions.GetCurrentUserName
</div>

Presumably, the primary use case for this is to get a base model to the view for all (or the majority of) controller actions.

Given that, I've used a combination of several of these answers, primary piggy backing on Colin Bacon's answer.

It is correct that this is still controller logic because we are populating a viewmodel to return to a view. Thus the correct place to put this is in the controller.

We want this to happen on all controllers because we use this for the layout page. I am using it for partial views that are rendered in the layout page.

We also still want the added benefit of a strongly typed ViewModel

Thus, I have created a BaseViewModel and BaseController. All ViewModels Controllers will inherit from BaseViewModel and BaseController respectively.

The code:

BaseController

public class BaseController : Controller
{
    protected override void OnActionExecuted(ActionExecutedContext filterContext)
    {
        base.OnActionExecuted(filterContext);

        var model = filterContext.Controller.ViewData.Model as BaseViewModel;

        model.AwesomeModelProperty = "Awesome Property Value";
        model.FooterModel = this.getFooterModel();
    }

    protected FooterModel getFooterModel()
    {
        FooterModel model = new FooterModel();
        model.FooterModelProperty = "OMG Becky!!! Another Awesome Property!";
    }
}

Note the use of OnActionExecuted as taken from this SO post

HomeController

public class HomeController : BaseController
{
    public ActionResult Index(string id)
    {
        HomeIndexModel model = new HomeIndexModel();

        // populate HomeIndexModel ...

        return View(model);
    }
}

BaseViewModel

public class BaseViewModel
{
    public string AwesomeModelProperty { get; set; }
    public FooterModel FooterModel { get; set; }
}

HomeViewModel

public class HomeIndexModel : BaseViewModel
{

    public string FirstName { get; set; }

    // other awesome properties
}

FooterModel

public class FooterModel
{
    public string FooterModelProperty { get; set; }
}

Layout.cshtml

@model WebSite.Models.BaseViewModel
<!DOCTYPE html>
<html>
<head>
    < ... meta tags and styles and whatnot ... >
</head>
<body>
    <header>
        @{ Html.RenderPartial("_Nav", Model.FooterModel.FooterModelProperty);}
    </header>

    <main>
        <div class="container">
            @RenderBody()
        </div>

        @{ Html.RenderPartial("_AnotherPartial", Model); }
        @{ Html.RenderPartial("_Contact"); }
    </main>

    <footer>
        @{ Html.RenderPartial("_Footer", Model.FooterModel); }
    </footer>

    < ... render scripts ... >

    @RenderSection("scripts", required: false)
</body>
</html>

_Nav.cshtml

@model string
<nav>
    <ul>
        <li>
            <a href="@Model" target="_blank">Mind Blown!</a>
        </li>
    </ul>
</nav>

Hopefully this helps.

Another option is to create a separate LayoutModel class with all the properties you will need in the layout, and then stuff an instance of this class into ViewBag. I use Controller.OnActionExecuting method to populate it. Then, at the start of layout you can pull this object back from ViewBag and continue to access this strongly typed object.