trying to put a strongly typed partial view on a homepage in asp.net but it wont seem to work it, here is my code
new to asp.net and partial views.
Controller :
public ActionResult VenuePartial()
{
ViewData["Message"] = _entities.VenuePartialList();
return View();
}
Repository :
public IEnumerable<Venue> VenuePartialList()
{
var list = from s in _entities.Venue
orderby s.name ascending
select s;
return list.ToList();
}
IRepository :
IEnumerable<Venue> VenuePartialList();
Index Page :
<%Html.RenderPartial("~/Views/Venue/VenuePartial.ascx");%>
Any help would be grateful asap please
regards T
Maybe you need to pass a model to this partial:
<% Html.RenderPartial("~/Views/Venue/VenuePartial.ascx", ViewData["Message"]); %>
And by the way WTF are you using ViewData["Message"] to pass a model instead of using a model and a strongly typed view:
public ActionResult VenuePartial()
{
return View(_entities.VenuePartialList());
}
and then:
<% Html.RenderPartial("~/Views/Venue/VenuePartial.ascx", Model); %>
This obviously assumes that your partial is strongly typed to IEnumerable<Venue>. If it is typed to a single Venue you might also consider using Editor/Display Templates. So in your main view:
<%= Html.DisplayForModel() %>
and in the corresponding display template (~/Views/Shared/DisplayTemplates/Venue.ascx):
<%# Control
Language="C#"
Inherits="System.Web.Mvc.ViewUserControl<AppName.Models.Venue>" %>
<span>
<%= Html.DisplayFor(x => x.SomePropertyOfVenue) %>
</span>
and now the display template will be rendered for each item of the model collection.
I have feature on my website (some UI and associated functionality) that I want to be able to reuse on multiple pages. For the purposes of this question, let's say it's a "Comments" feature.
There is an area in my application for Components and within the area are a controller: /Controllers/CommentController, and two partial views: /Views/Comment/Index.ascx (for listing comments) and /Views/Comment/Create.ascx (for creating comments).
CommentController looks something like this:
public class CommentController : Controller
{
[ChildActionOnly]
public ActionResult Index()
{
return PartialView(GetComments());
}
[HttpGet]
[ChildActionOnly]
public ActionResult Create()
{
return PartialView(); //this is wrong.
}
[HttpPost]
[ChildActionOnly]
public ActionResult Create(FormCollection formValues)
{
SaveComment(formValues);
return RedirectToAction("Index"); //this is wrong too.
}
}
Index Partial View:
<%# Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<dynamic>" %>
<div>
<% foreach (var item in Model) { %>
<div>
<%: item.Comment %>
</div>
<% } %>
<%: Html.ActionLink("Add a Comment", "Create", "Comment", new { area = "Components" }, null) %>
</div>
Create Partial View:
<%# Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<dynamic>" %>
<div>
<% using (Html.BeginForm())
{%>
Enter your comment:
<div>
<input type="text" name="comment" />
</div>
<p>
<input type="submit" value="Create" />
<% //also render a cancel button and redirect to "Index" view %>
</p>
<% } %>
</div>
The Index partial view is included in a view with RenderAction, like so:
<% Html.RenderAction("Index", "Comment", new { area = "Components" }); %>
This code doesn't work because the forms within the partial views submit to actions on the CommentsController that are marked [ChildActionOnly] (this is by design, I don't want the "Components" to be requested independently of a hosting page).
How can I make this "component" approach work, i.e. have a partial view that submits a form to change the state of a component within a page without losing the hosting page itself?
EDIT:
To clarify, the use of [ChildActionOnly] is not my problem here. If I remove the attribute from my action methods, my code only "works" in that it doesn't throw an exception. My "component" still breaks out of its hosting page when its form is submitted (because I'm telling the form to submit to the partial view's URL!).
You are making MVC fight itself by asking a form to target an action that is marked as ChildActionOnly.
My solution to this problem when I was designing a highly reusable wizard framework, was to NOT mark the actions as ChildActionOnly but instead to detect if the request was an ajax one or just a plain vanilla request.
The code for all this is packaged into a base controller class. In your derived controllers, you do something like:
[WizardStep(4, "Illness Details")]
public ActionResult IllnessDetails()
{
return Navigate();
}
Where the Navigate() method of the base controller has decided whether to return the full view or just the partial view, depending on whether it is, or isn't, an ajax request. That way, you can never return the partial view in isolation.
To ascertain if it is an Ajax request, I used a combination of Request.IsAjaxRequest() and TempData. The TempData is needed because my wizard framework implements the PRG pattern out of the box, so I need to persist the fact that the original post was an ajax one.
I guess this is just one solution and it took a bit of trial and error to get it right. But now I live happily ever after developing wizards like I was JK Rowling...
Use Ajax to post the partial.
After so many years using ASP.Net, I’m still trying to figure out how to achieve the same results using MVC.
I have a materpage with a control that is strongly type to something. When I navigate to a view of a different strongly type model ...and click on the button to execute something, I get "The model item passed into the dictionary is of type Site.Models.RegisterModel', but this dictionary requires a model item of type Site.Models.LogOnModel'".
For the sake of this example, we can take the Default MVC app that is provided with VS 2010, let’s imagine I want to change the “LogonUserControl.ascx” so that it either tells me the logged user (as it works currently) OR allow me to login from there, showing me the text boxes for username and password (therefore in this case from the home page).
So I take the control and strongly type it as:
<%# Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<Gioby.Models.LogOnModel>" %>
<%
if (Request.IsAuthenticated) {
%>
Welcome <b><%: Page.User.Identity.Name %></b>
[ <%: Html.ActionLink("Log Off", "LogOff", "Account")%> ]
<%
}
else {
%>
<% using (Html.BeginForm()) { %>
<div id="logon">
<div class="editor-label">
<%: Html.LabelFor(m => m.UserName)%>
<%: Html.TextBoxFor(m => m.UserName)%>
<%: Html.ValidationMessageFor(m => m.UserName, "*") %>
<%: Html.LabelFor(m => m.Password)%>
<%: Html.PasswordFor(m => m.Password)%>
<%: Html.ValidationMessageFor(m => m.Password, "*") %>
<input type="submit" value="Log On" />
</div>
<div class="editor-label">
<%: Html.ActionLink("Register here", "Register", "Account")%>
<%: Html.CheckBoxFor(m => m.RememberMe, new { #class = "pad-left" })%>
<%: Html.LabelFor(m => m.RememberMe) %>
</div>
</div>
<% } %>
<%
}
%>
Then on the HomeController, I add a procedure as:
[HttpPost]
public ActionResult Index(LogOnModel model, string returnUrl)
{
if (ModelState.IsValid)
{
// ==>> Check Login against your DB
// Now check if param returnUrl is empty
if (!String.IsNullOrEmpty(returnUrl))
return Redirect(returnUrl);
return RedirectToAction("Index", "Home");
}
// If we got this far, something failed, redisplay form
return View(model);
}
I tested it from the home page … it works !!!
BUT when I navigate to the “Register” view (remember that the “LogonUserControl.ascx” is located inside the “MasterPage”, therefore visible from the Register view).
So when I click on the Register button, I get the error:
The model item passed into the dictionary is of type Site.Models.RegisterModel', but this dictionary requires a model item of type Site.Models.LogOnModel'.
QUESTION:
Does that mean that I will never be able to different pieces together into one view?
Let’s say I want to write an eCommerce site and on the home page I want to see “Most used Tags”, “Most bought products”, “Product of the Month”, “List of Categories” …all within the same view and each one with his own HTTP POST action.
If this is possible using MVC?
If I'm understanding the problem correctly, you have two Views that use the same MasterPage, but which are strongly typed against different ViewModels. The master page is able to include a Partial View that is also strongly typed, as long as its expected ViewModel is the same as that of the parent view. However, if you're using a view with a different ViewModel type, it doesn't know what to do.
Consider the following:
<% Html.RenderPartial("LogOn") %>
The above code implicitly includes the model data for the current view being rendered. It's exactly the same as if you had said:
<% Html.RenderPartial("LogOn", Model) %>
So this will only work if Model is a LogOnModel. Remember that the MasterPage is really a part of whatever View inherits it, so even if you're putting this in the MasterPage, it's as if you'd put the same code in every view that inherits it. So if your View's Model is not the same as the PartialView's Model, this won't work. One alternative is to use inheritance to ensure that every ViewModel will include all the information required by the Master Page. This approach is described in detail here.
But that approach means that you have to always use a factory to produce your view model, and every view model has to be somewhat aware of which master page it will use. In our product, we can use a different master page on the same view depending on what mode the user is viewing the site in, so it doesn't make sense to tie the ViewModel to that of the Master Page. We accomplish what you're describing using the RenderAction method, which allows you to render an entire controller action as if it were just a part of the larger view. Some of the advantages of this approach are discussed here.
So now you can have your MasterPage include whatever little partial views you want, but you separate the logic for building the ViewModel of each of these Views into an individual controller action that's responsible for that particular Partial View:
<% Html.RenderAction("LogOnBox") %>
The Action:
public ActionResult LogOnBox()
{
LogOnModel model = GetLogOnModel();
return PartialView("LogOnUserControl", model);
}
Now, regardless of what model your current view uses, your Master Page can include “Most used Tags”, “Most bought products”, “Product of the Month”, “List of Categories”, etc. Better still, these portions of the page can leverage output caching so they don't have to be regenerated with every page load if they don't change very often.
I am getting my content from a database. How can i use partial views to show content on page using that database?
database table: Content
[Id, Content] these are 2 fields
i want to get the content from db using partial views.
How i will pass an id to a partial view and show the content in view page?
You could use Html.RenderAction:
public class MyController
{
[ChildActionOnly]
public ActionResult Foo(int id)
{
var content = GetContentFromDatabase(id);
return Content(content, MediaTypeNames.Text.Html);
}
}
And in your view include the partial:
<%= Html.RenderAction("foo", "mycontroller", new { id = 5 }) %>
Remark: RenderAction is part of the now released ASP.NET MVC 2 RTM. For ASP.NET MVC 1 you may take a look at the Futures assembly containing this extension method.
Inside your view, use the Html.RenderPartial function. There are a few different uses:
You can pass in a model to the partial view: <% Html.RenderPartial("partialName", model); %>
Or you can pass in a whole new ViewDataDictionary: <% Html.RenderPartial("partialName", viewData); %>
For the full documentation, see here.
EDIT: (Answer to comment):
I would include that data as part of you're view's model. For example, let's say in your model you have:
List<Person> People;
In your view, you want to loop through each one of these, and use a PartialView to display the details:
<% foreach( var p in Model.People){ %>
<p> <% Html.RenderPartial("personPartial", p); %> </p>
<%}%>
Now, your PartialView might look like:
<%# Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<Person>" %>
<%=Model.PersonName%>
Now I've seen some questions like this, but it's not exactly what I want to ask, so for all those screaming duplicate, I apologize :).
I've barely touched ASP.NET MVC but from what I understand there is no ViewState/ControlState... fine. So my question is what is the alternative to retaining a control's state? Do we go back to old school ASP where we might simulate what ASP.NET ViewState/ControlState does by creating hidden form inputs with the control's state, or with MVC, do we just assume AJAX always and retain all state client-side and make AJAX calls to update?
This question has some answers, Maintaining viewstate in Asp.net mvc?, but not exactly what I'm looking for in an answer.
UPDATE: Thanks for all the answers so far. Just to clear up what I'm not looking for and what I'm looking for:
Not looking for:
Session solution
Cookie solution
Not looking to mimic WebForms in MVC
What I am/was looking for:
A method that only retains the state on postback if data is not rebound to a control. Think WebForms with the scenario of only binding a grid on the initial page load, i.e. only rebinding the data when necessary. As I mentioned, I'm not trying to mimic WebForms, just wondering what mechanisms MVC offers.
The convention is already available without jumping through too many hoops. The trick is to wire up the TextBox values based off of the model you pass into the view.
[AcceptVerbs(HttpVerbs.Get)]
public ActionResult CreatePost()
{
return View();
}
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult CreatePost(FormCollection formCollection)
{
try
{
// do your logic here
// maybe u want to stop and return the form
return View(formCollection);
}
catch
{
// this will pass the collection back to the ViewEngine
return View(formCollection);
}
}
What happens next is the ViewEngine takes the formCollection and matches the keys within the collection with the ID names/values you have in your view, using the Html helpers. For example:
<div id="content">
<% using (Html.BeginForm()) { %>
Enter the Post Title: <%= Html.TextBox("Title", Model["Title"], 50) %><br />
Enter the Post Body: <%= Html.TextArea("Body", Model["Body"]) %><br />
<%= Html.SubmitButton() %>
<% } %>
</div>
Notice the textbox and textarea has the IDs of Title and Body? Now, notice how I am setting the values from the View's Model object? Since you passed in a FormCollection (and you should set the view to be strongly typed with a FormCollection), you can now access it. Or, without strongly-typing, you can simply use ViewData["Title"] (I think).
POOF Your magical ViewState. This concept is called convention over configuration.
Now, the above code is in its simplest, rawest form using FormCollection. Things get interesting when you start using ViewModels, instead of the FormCollection. You can start to add your own validation of your Models/ViewModels and have the controller bubble up the custom validation errors automatically. That's an answer for another day though.
I would suggest using a PostFormViewModel instead of the Post object, but to each-his-own. Either way, by requiring an object on the action method, you now get an IsValid() method you can call.
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult CreatePost(Post post)
{
// errors should already be in the collection here
if (false == ModelState.IsValid())
return View(post);
try
{
// do your logic here
// maybe u want to stop and return the form
return View(post);
}
catch
{
// this will pass the collection back to the ViewEngine
return View(post);
}
}
And your Strongly-Typed view would need to be tweaked:
<div id="content">
<% using (Html.BeginForm()) { %>
Enter the Post Title: <%= Html.TextBox("Title", Model.Title, 50) %><br />
Enter the Post Body: <%= Html.TextArea("Body", Model.Body) %><br />
<%= Html.SubmitButton() %>
<% } %>
</div>
You can take it a step further and display the errors as well in the view, directly from the ModelState that you set in the controller.
<div id="content">
<%= Html.ValidationSummary() %>
<% using (Html.BeginForm()) { %>
Enter the Post Title:
<%= Html.TextBox("Title", Model.Title, 50) %>
<%= Html.ValidationMessage("Title") %><br />
Enter the Post Body:
<%= Html.TextArea("Body", Model.Body) %>
<%= Html.ValidationMessage("Body") %><br />
<%= Html.SubmitButton() %>
<% } %>
</div>
What is interesting with this approach is that you will notice I am not setting the validation summary, nor the individual validation messages in the View. I like to practice DDD concepts, which means my validation messages (and summaries) are controlled in my domain and get passed up in the form of a collection. Then, I loop throught he collection (if any errors exist) and add them to the current ModelState.AddErrors collection. The rest is automatic when you return View(post).
Lots of lots of convention is out. A few books I highly recommend that cover these patterns in much more detail are:
Professional ASP.NET MVC 1.0
Pro ASP.NET MVC 1.0 Framework
And in that order the first covers the raw nuts and bolts of the entire MVC framework. The latter covers advanced techniques outside of the Microsoft official relm, with several external tools to make your life much easier (Castle Windsor, Moq, etc).
The View is supposed to be dumb in the MVC pattern, just displaying what the Controller gives it (obviously we do often end up with some logic there but the premise is for it not to be) as a result, controls aren't responsible for their state, it'll come from the controller every time.
I can't recommend Steven Sanderson's book Pro ASP.NET MVC by Apress enough for getting to grips with this pattern and this implementation of it.
In Web Forms, control values are maintained in the viewstate so you (theoretically) don't need to reinitialize and such with each postback. The values are (again theoretically) maintained by the framework.
In ASP.NET MVC, if you follow the paradigm, you don't need to maintain state on form elements. The form element values are available on post where your controller can act on them (validation, database updates, etc.). For any form elements that are displayed once the post is processed, you (the developer) are responsible for initializing them - the framework doesn't automatically do that for you.
That said, I have read about a mechanism called TempData that allows your controller to pass data to another controller following a redirect. It is actually a session variable (or cookie if you configure it as such) but it is automatically cleaned up after the next request.
The answer really depends on the types of controls you are trying to maintain state for. For basic Html controls then it is very easy to maintain state with your Models, to do this you need to create a strongly typed view.
So if we had a User model with the properties: Username, FullName, Email, we can do the following in the view:
<%= Html.ValidationSummary() %>
<% using (Html.BeginForm()) { %>
<fieldset>
<legend>User details</legend>
<%= Html.AntiForgeryToken() %>
<p>
<label for="Username">Username:</label>
<%= Html.Textbox("Username", Model.Username, "*") %>
</p>
<p>
<label for="FullName">FullName:</label>
<%= Html.Textbox("FullName", Model.FullName, "*") %>
</p>
<p>
<label for="Email">Email:</label>
<%= Html.Textbox("Email", Model.Email, "*") %>
</p>
<p>
<input type+"submit" value="Save user" />
</p>
</fieldset>
<% } %>
We would then have two controller actions that display this view, one for get and another for post:
[AcceptVerbs(HttpVerbs.Get)]
public ActionResult User()
{
return View(new User())
}
[AcceptVerbs(HttpVerbs.Post)]
[ValidateAntiForgeryToken]
public ActionResult User([Bind(Include = "Username,FullName,Email")]User user)
{
if (!ModelState.IsValid()) return View(user);
try
{
user.save()
// return the view again or redirect the user to another page
}
catch(Exception e)
{
ViewData["Message"] = e.Message;
return View(user)
}
}
Is this what you are looking for? Or do you want to maintain the state of Models that are not being displayed in a form between requests?
The key thing to remember is that your code executes on the server for the duration of the request and ends, the only information you can pass between your requests is basic html form data, url parameters and session information.
As other people have mentioned, I'd highly recommend Steve Sandersan's Pro ASP.NET MVC Framework for a complete understanding of working with the MVC Framework.
hidden fields, like:
<% using (Html.BeginForm<SomeController>(c=>c.SomeAction(null))) {%>
<%= Html.Hidden("SomeField", Model.SomeField)%>
<%= Html.Hidden("AnotherField", Model.AnotherField)%>
setting the specific model & not having any explicit fields (gives u hidden fields). In the example below, the Model is filled by the controller with values received from the last post, so this enables a no js option in the page that can filter based on a status:
Some Filter: <% using( Html.BeginForm<SomeController>(
c => c.SomeAction(model.SomeField, model.AnotherField, model.YetAnotherField, null, model.SomeOtherField)
)) { %>
<%= Html.DropDownList("status", Model.StatusSelectList)%>
<input type="submit" value="Filter" class="button" />
<% } %>
use extension methods to create fields, if u just want the fields to be filled with posted values when u are showing failed validation messages on the submitted form
on asp.net mvc 2 they introduced a way to save an instance in a hidden field ... encoded + (I think) signed
TempData if everything of the above doesn't do it (goes through session - cleaned on the next request)
as u mentioned, when using ajax the state is already in the previously loaded fields in the client site. If u l8r on need to do a full post, update any field u might need to with your js.
The above are all different independent options to achieve it that can be used in different scenarios. There are more options I didn't mention i.e. cookies, session, store stuff in db (like for a resumable multi step wizard), parameters passed to an action. There is no 1 single mechanism to rule them all, and there shouldn't be.
The best way to do this, i think, is to serialize your original model to a hidden field, then deserialize it and update the model on post. This is somewhat similair to the viewstate approach, only you have to implement it yourself. I use this:
first i need some methods that make things easier:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Web.Mvc;
using LuvDaSun.Extensions;
using System.Web.UI;
namespace LuvDaSun.Web.Mvc
{
public static class HtmlHelperExtensions
{
static LosFormatter _losFormatter = new LosFormatter();
public static string Serialize(this HtmlHelper helper, object objectInstance)
{
var sb = new StringBuilder();
using (var writer = new System.IO.StringWriter(sb))
{
_losFormatter.Serialize(writer, objectInstance);
}
return sb.ToString();
}
}
[AttributeUsage(AttributeTargets.Parameter)]
public class DeserializeAttribute : CustomModelBinderAttribute
{
public override IModelBinder GetBinder()
{
return new DeserializeModelBinder();
}
}
public class DeserializeModelBinder : IModelBinder
{
static LosFormatter _losFormatter = new LosFormatter();
public object BindModel(ControllerContext controllerContext, ModelBindingContext bindingContext)
{
if (bindingContext.ModelType.IsArray)
{
var type = bindingContext.ModelType.GetElementType();
var serializedObjects = (string[])bindingContext.ValueProvider.GetValue(bindingContext.ModelName).ConvertTo(typeof(string[]));
var deserializedObjects = Array.CreateInstance(bindingContext.ModelType.GetElementType(), serializedObjects.Length);
for (var index = 0; index < serializedObjects.Length; index++)
{
var serializedObject = serializedObjects[index];
var deserializedObject = _losFormatter.Deserialize(serializedObject);
deserializedObjects.SetValue(deserializedObject, index);
}
return deserializedObjects;
}
else
{
var serializedObject = (string)bindingContext.ValueProvider.GetValue(bindingContext.ModelName).ConvertTo(typeof(string));
var deserializedObject = _losFormatter.Deserialize(serializedObject);
return deserializedObject;
}
}
}
}
then in my controller i have something like this (to update a product)
public ActionResult Update(string productKey)
{
var model = _shopping.RetrieveProduct(productKey);
return View(model);
}
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult Update([Deserialize]Shopping.IProduct _model, FormCollection collection)
{
UpdateModel(model);
model.Save();
return RedirectAfterPost();
}
and i need a hidden field that holds the serialized object in the form:
<%
using (Html.BeginRouteForm("Product", FormMethod.Post, new { id = UniqueID, }))
{
%>
<%= Html.Hidden("Model", Html.Serialize(Model)) %>
<h1>
Product bewerken</h1>
<p>
<label for="<%=UniqueID %>_Name">
Naam:</label>
<input id="<%=UniqueID %>_Name" name="Name" type="text" value="<%= Html.AttributeEncode(Model.Name) %>"
class="required" />
<br />
</p>
<p>
Omschrijving:<br />
<textarea id="<%= UniqueID %>_Description" name="Description" cols="40" rows="8"><%= Html.Encode(Model.Description) %></textarea>
<br />
</p>
<p>
<label for="<%=UniqueID %>_Price">
Prijs:</label>
<input id="<%= UniqueID %>_Price" name="Price" type="text" value="<%= Model.Price.ToString("0.00") %>"
class="required" />
<br />
</p>
<ul class="Commands">
<li>Annuleren</li>
<li>
<input type="submit" value="Opslaan" /></li>
</ul>
<%
}
%>
<script type="text/javascript">
jQuery('#<%= UniqueID %>').validate();
</script>
as you can see, a hidden field (Model) is added to the form. It contains the serialization information for the original object. When the form is posted, the hidden field is also posted (ofcourse) and the contents are deserialized by the custom modelbinder to the original object which is then updated and saved by the controller.
Do note that the object you are serializing needs to be decorated with the Serializable attribute or needs to have a TypeConverter that can convert the object to a string.
The LosFormatter (Limited Object Serialization) is used by the viewstate in webforms. It also offers encryptionn of the serialization data.
greets...
AJAX calls is what we do. If you're talking about grids in general, check out JQGrid and how they recommend the AJAX implementation.