I am looking to rewrite an existing ASP.NET MVC application to .Net 5 Razor pages and one of the issues I am struggling to resolve is how to create a partial razor page that creates it's own model rather than having the model passed in from the parent page.
In MVC, I would simply add an action to the controller and specify this within the view;
#Html.Action("_SideNavigation", "Shared")
Ultimately, this would call the _SideNavigation action on the share controller, which would then create the model and pass it back to the view.
I am trying to achieve something similar in Razor Pages, I do not want to define the model of the partial within the model of the parent view, such as this;
<partial name="_Partial1" model="Model.PartialModel" />
I have tried creating an empty model within the parent page and assumed this would then call the OnGet() method within the partial itself;
<partial name="_Test" model="new _TestModel()" />
public class _TestModel : PageModel
{
public TopNavigation TopNavigation { get; set; } = new TopNavigation();
public void OnGet()
{
TopNavigation = new TopNavigation { NotificationCount = 1 };
}
}
and then referencing the model in the page;
#Model.TopNavigation.NotificationCount
But it is always null.
Is this behaviour not available in razor pages?
Can someone help me understand partial views, forms and posting in ASP.NET Core Razor.
I have a Search.cshtml partial view located in "~/Client/Search" :
#model Web.Pages.Client.SearchModel
#using (Html.BeginForm())
{
<div>
#Html.RadioButtonFor(x => x.searchType, (int)ApplicationCore.Interfaces.SearchType.mobile, new { Name = "SearchType" }) Mobile
#Html.RadioButtonFor(x => x.searchType, (int)ApplicationCore.Interfaces.SearchType.phone, new { Name = "SearchType" }) Phone
#Html.RadioButtonFor(x => x.searchType, (int)ApplicationCore.Interfaces.SearchType.email, new { Name = "SearchType" }) Email
</div>
#Html.TextBoxFor(x => x.searchFilter)
<input type="submit" value="Search"/>
}
With code page Search.cshtml.cs :
public class SearchModel : PageModel
{
public SearchType searchType { get; set; }
public string searchFilter { get; set; }
private readonly IClientService _clientService;
private readonly Infrastructure.Data.DBContext _context;
public SearchModel(Infrastructure.Data.DBContext context, IClientService clientService)
{
_context = context;
_clientService = clientService;
searchFilter = string.Empty;
searchType = SearchType.mobile;
}
public async Task<IActionResult> OnPostAsync()
{
return RedirectToPage("./Index");
}
}
If I load the "~/Client/Search" Partial View directly it loads and on post it correctly fires the OnPosAsync() action.
However if the "~/Client/Search" Partial View is rendered from the "~/Session/CheckIn" parent View :
#await Html.PartialAsync("~/Client/Search", Model._searchModel)
The OnPostAsync() within the "~/Client/Search" Partial View no longer fires.
I have tried all sorts of combinations to define "action", "controller" within the Html.BeginForm in the Partial View, however I can never get the OnPostAsync() within the Partial View to fire.
Any pointers? Read a lot of articles and forum posts however there are no clear descriptions or walkthroughs to help me understand this and get the Partial View action method firing on postback from parent View.
This is why Razor Pages are a blight: they obfuscate logic, allowing people to build stuff without ever actually understanding how any of it works. </rant>
There's nothing special about a partial view. It's just a view like any other view. What makes it "partial" is the context in which it's used, i.e. injecting it into the rendering of a view. As such, Razor Pages lets you add a code-behind, because it's just a view, and any view can have a Razor Pages code-behind. However, when used like a partial, that code-behind is not actually utilized, and there's your problem.
Also, you need to bear in mind that the whole concept of partial views only exists server-side. Once the response has been returned, all you have is just an HTML document. The browser couldn't care less whether you used one partial view, 100 partial views or no partial views to create the response server-side. As such, using a partial doesn't somehow magically buy you the ability to just work with a single section of your page, such that when you post, only that section is changed. For that, you need AJAX. Otherwise, doing a post, whether from a "partial view" or not, will cause the entire view to be changed in the browser window.
In other words, you need something server-side that will respond to that post request by returning a new full view, not just your partial, or you need to make the request client-side via AJAX, and return just your partial view. However, then, you're responsible for replacing whatever HTML should be replaced with that response, yourself.
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?
If you are required to pass the same properties to each page, then creating a base viewmodel that is used by all your view models would be wise. Your layout page can then take this base model.
If there is logic required behind this data, then this should be put into a base controller that is used by all your controllers.
There are a lot of things you could do, the important approach being not to repeat the same code in multiple places.
Edit: Update from comments below
Here is a simple example to demonstrate the concept.
Create a base view model that all view models will inherit from.
public abstract class ViewModelBase
{
public string Name { get; set; }
}
public class HomeViewModel : ViewModelBase
{
}
Your layout page can take this as it's model.
#model ViewModelBase
<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="width=device-width" />
<title>Test</title>
</head>
<body>
<header>
Hello #Model.Name
</header>
<div>
#this.RenderBody()
</div>
</body>
</html>
Finally set the data in the action method.
public class HomeController
{
public ActionResult Index()
{
return this.View(new HomeViewModel { Name = "Bacon" });
}
}
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.
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.
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>
Mind Blown!
</li>
</ul>
</nav>
Hopefully this helps.
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.
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>();
}
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.
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)
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.
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
The best way to use static strings such as page title, page name and the location etc, is to define via ViewData. Just define required ViewData in ViewStart.cshtml
#{
Layout = "_Layout";
ViewData["Title"] = "Title";
ViewData["Address"] = "1425 Lane, Skardu,<br> Pakistan";
}
and call whenever require like
<div class="rn-info-content">
<h2 class="rn-contact-title">Address</h2>
<address>
#Html.Raw(ViewData["Address"].ToString())
</address>
</div>
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>
In .NET Core, you can use View Components to do this.
https://learn.microsoft.com/en-us/aspnet/core/mvc/views/view-components?view=aspnetcore-5.0
From the link above, add a class Inheriting from ViewComponent
using Microsoft.AspNetCore.Mvc;
using Microsoft.EntityFrameworkCore;
using System.Collections.Generic;
using System.Linq;
using System.Threading.Tasks;
using ViewComponentSample.Models;
namespace ViewComponentSample.ViewComponents
{
public class PriorityListViewComponent : ViewComponent
{
private readonly ToDoContext db;
public PriorityListViewComponent(ToDoContext context)
{
db = context;
}
public async Task<IViewComponentResult> InvokeAsync(
int maxPriority, bool isDone)
{
var items = await GetItemsAsync(maxPriority, isDone);
return View(items);
}
private Task<List<TodoItem>> GetItemsAsync(int maxPriority, bool isDone)
{
return db.ToDo.Where(x => x.IsDone == isDone &&
x.Priority <= maxPriority).ToListAsync();
}
}
}
Then in your view (_layout in my case)
#await Component.InvokeAsync("PriorityList", new { maxPriority = 4, isDone = true })
If you need a view, make a folder at ~/Views/Shared/Components/<Component Name>/Default.cshtml. You need to make the folder Components then in that, make a folder with your component name. In the example above, PriorityList.
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
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".
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)
{
#item.Name
}
</div>
</div>
</div>
</div>
I am a beginner in ASP.NET MVC.
My page has one partial view called _Navigation that I am reusing.
If the user is in the "Home" the <a> of the navigation needs to point to the "#" char, if the user is in the "Services" page, the href of the navigation needs to point to other url, let's say "www.mysite.com". It will occur with other links in this menu too.
I tried to do the following
#if (ViewContext.RouteData.Values.ContainsValue("Services"))
{
#model MySite.Models.ServicesNavigation
}
else
{
#model MySite.Models.HomeNavigation
}
But it says I can have only one model.
How to solve it?
You can try using Interface.
public interface INavigation
{
//Your props here
}
public class ServicesNavigation : INavigation
{
}
public class HomeNavigation: INavigation
{
}
Then your view can be of type INavigation.
#model INavigation
And in your controller based on your conditions you can pass the impementation of INavigation you want.
.......
INavigation model;
if(conditionOneIsMet)
{
model = new ServicesNavigation();
}
else
{
model = new HomeNavigation();
}
return View(model);
Your view is in fact a class derived from the WebViewPage<TModel> class. The #model statement defines type of the model (TModel) Because it is the compile time statement, you can't change it in run time.
If you need two different models, you should have two different views.
At the end of my view, I call this:
<%= Html.Action("ProductLayoutAndLimits", this.Model) /* Render product-specific options*/ %>
That action is virtual in my controller:
[ChildActionOnly]
public virtual ActionResult ProductLayoutAndLimits(DeliveryOptionsViewModel optionsViewModel)
{
return new EmptyResult();
}
The intent was that I would override this method in a product specific controller. So naturally, I did this:
public override System.Web.Mvc.ActionResult ProductLayoutAndLimits(DeliveryOptionsViewModel optionsViewModel)
{
But the breakpoint isn't hitting, so my override is not getting picked up. Is there a different overload I should be using? Or do I need to pass in a different object? Or is there an annotation that I need on my product specific action in order for it to be detected?
Any help is appreciated. Thanks!
Edit
While all suggestions are appreciated, I am most interested in solutions that actually answer my question, rather than suggesting a different technique.
Templates have been suggested, but please note that I need controller code to be executed before any new additional view code is rendered. The base controller is in a solution that serves as a platform to other products. They cannot do anything product specific. After they render their view, the intent is that my override of the child action will be called. My controller code will check a number of things in order to determine how to set properties on my model before it renders the view.
Edit
I found the problem. I feel silly, as usual. Html.Action was being called from Platform's view code. It turned out that we have been using a product specific view for this since July. I didn't notice because we don't typically use product specific views. Whoops!
Why are you using child actions and overriding controllers to handle this? Why not display templates:
<%= Html.DisplayForModel("ProductLayoutAndLimits") %>
where ProductLayoutAndLimits would be the name of the corresponding display template.
Example:
Model:
public class BaseViewModel
{
public string BaseProp { get; set; }
}
public class DerviedViewModel : BaseViewModel
{
public string DerivedProp { get; set; }
}
Controller:
public class HomeController : Controller
{
public ActionResult Index()
{
return View(new DerviedViewModel
{
BaseProp = "base prop",
DerivedProp = "derived prop"
});
}
}
View (~/Views/Home/Index.cshtml):
#model AppName.Models.BaseViewModel
#Html.DisplayForModel()
Display Template (~/Views/Home/DisplayTemplates/DerviedViewModel.cshtml):
#model AppName.Models.DerviedViewModel
#Html.DisplayFor(x => x.BaseProp)
#Html.DisplayFor(x => x.DerivedProp)
Of course you could also have a display template for the base class (~/Views/Home/DisplayTemplates/BaseViewModel.cshtml):
#model AppName.Models.BaseViewModel
#Html.DisplayFor(x => x.BaseProp)
and this template would have been rendered if your controller action had returned the base class as model.