MVC :: How To Display Property Of Model On View
Aug 26, 2010How can I show display property of Model on View.
public class Model
{
[Required]
[DataType(DataType.Text)]
[DisplayName("First Name")]
[code]...
How can I show display property of Model on View.
public class Model
{
[Required]
[DataType(DataType.Text)]
[DisplayName("First Name")]
[code]...
but I am new with MVC and database-driven applications.
I hava a table with Cities (ID,Name) and one with Businesses(ID,CityID,Name). The later has a foreign key relationship with the first table.
I used a linq-to-SQL class to create my models.
My problem is that I need to access the City Name(not the ID) when displaying the Business Model View.
How can I do this? if from my controller I call the View passing the auto-generated Business Model, I only have access to the ID.
So I have code as below
public ActionResult Create(AccountClassificationCreateViewModel ACCVM)
{
if (ACCVM.newMtn0130 == null)//Why ACCVM.newMtn0130 is null after [code]...
I wonder why the ACCVM.newMtn0130 become null after RedirectToAction? I already assign with "_ACCVM.newMtn0130 = newMtn0130".
ACCVM._Error didn't become null after RedirectToAction.
Curious what the best practice is for returning errors to a view from a controller where the error isn't really a validation error, but more like "user not found" or "service timeout" type of errors. (in ASP.NET MVC2 framework)
I've been adding them to the ModelState's model errors, but that doesn't seem appropriate. (although easy to implement and maintain)
Example - A user tries to log in, and their credentials do not match a known user.
I have been reading about different model for development
model view control mvc
model view presenter MVP
Model view view model MVVM
i belive MVC has two big Advantage over webform 1) TDD 2) More control on HTML
MVP is bit variation in mvc model. rapid development as well as 1) TDD 2) More control on HTML (correct me if i m wrong) see the below link
[URL]
I've got a view model that has some text properties. And I use Html.DisplayFor to display them on the screen. But those text data won't be post back, so in the case of failed validation, the returned view won't have those data. How do I handle this kind of situation?
View 2 RepliesI've been used to decorating data model classes with data annotation attributes, but the purist in me baulks slightly at including purely presentational attributes such as display format here. I am, however, quite happy to keep validation centric attributes here. One good reason I have to continue keeping all annotations etc. in the data model is that my view model aggregates data model classes, e.g.
my ViewModelBase.DetailItem<TEntity> property in the view model is just a reference to an entity class in my data model. If I wanted to move presentational annotations to the view model, I would have to quite radically revise my design to one where I duplicate data model properties in my view model and use an object mapping tool to populate view model objects based on data model objects.
[code]....
Say I got a domain model as follows: (and my repository expect an instance of this object)
[Code]....
And a view model (which my views are based on)
[Code]....
At the moment I got it like this and have my controller action manually create a new Person object from the PersonModel object before passing it on to the repository, which does not feel right.
So I tried to have PersonModel inherit from Person and pass that to the repository (also tried casting the PersonModel to a Person object first), but that don't work out.
What's the right way to have PersonModel automatically cast to Person? I want to keep this logic as my current structure allow me to keep things really loosely coupled, with the repository layer not having a clue how it's being used.
I have an entity object as model input in my view. Not sure if it's the best approach, but I use the object references to get values from a related object.Let's say I've got a car entity and a manufacturer entity.Here's how it would look in my view
[Code]....
I've get a nullreferenceexception where the car entitity does not have a reference towards manufacturer. I'd like to output a " " if no manufacturer exists.
've created a Search partial view and it works, except for my view model. What i mean is, the partial view sends a string to my ActionResult Method, but only a string, that is not related to the view model that the partial view is built in. If i tell my ActionResult method to receive the view model in which the partial view is built in, it always tells me that i have null values. Here's my code This is my partial view:
[Code]....
Now here is my Model in my MoviesSearch view model:
[Code]....
Now the actual MoviesSearchViewModel
[Code]....
Inside my Search controller i have this
[Code]....
And then i have this:
one thing that has been puzzling me since learning MVC2 is the following case scenario: I have a view which contains two latest news lists, a login form and a signup form. Every example I found on Views and View Models so far has a one-to-one example such as a simple login form etc. But how do I create a model that provides the properties and validation for a login and signup form and manages the data for the news lists. can I pass multiple models in the strongly typed view? When I created one model the form validation would fail as it expects all fields - login and signup to be filled. I am missing some advanced examples or information.
View 2 Repliesmaking my first steps with asp.net mvc and I actually create a (very) small website.I have 1 controller : TasksController1 view : Tasks/Index.aspx1 Partial View : Tasks/AvailableSorting.ascx In my controller I have 2 methods :
[Code]....
In my Tasks/Index.aspx, I add my partial view (<% Html.RenderPartial("AvaliableSorting"); %>)
That sounds simple. But, It just doesn't work : the Index return a list of Task and the parial view (should) return a list of string.The index.aspx works perfectly, but the partial view doesn't get the right model ! he gets the model of the Index page (list of Task).
I would like to add a partial view to my view page
The view page contain a mode name "<IEnumerable<JvTake1.Models.VortN>>
And the partial view contain a model name "<IEnumerable<JvTake1.Models.VideoL>>"
So, theoretically I think they were suppose to work together but no Can any one guide me how to make it happen?
I've got problem with my app .
I've got such classes (this is some kind of tree structure):
[Code]....
[Code]....
in Index() action i've got this piece of code
[Code]....
[Code]....
I am totally a new bee in asp.net mvc. Let me now explain my issue. I have a strongly typed view which inherits from a view model Document. I want to have a partial view in that view itself. My code for this purpose is :
<%@ Page Title="" Language="C#" MasterPageFile="~/Views/Shared/Site.Master" Inherits="System.Web.Mvc.ViewPage<HexsolveMVC.Models.Document>" %>
<asp:Content ID="Content1" ContentPlaceHolderID="TitleContent" runat="server">
DetailsDocumentTemplate
</asp:Content>
<asp:Content ID="Content2" ContentPlaceHolderID="MainContent" runat="server">
<div>
<div>
<div>
2.4</div>
<div>
Published</div>
<h1>
Document Title: <span>
<%=Model.DocTitle %></span><span><a href="#">[ Edit ]</a></span></h1>
<ul>
<li>
<label>
<span>*</span>Created Date:</label>
<span>
<%= Model.DocCreatedDate%></span></li>
<li>
<label>
Status:</label>
<span>.................................
I have a view that is not strongly typed. However I have in this view a partial view that is strongly typed.How do I do I pass the model to this strongly typed view?I tried something like
public ActionResult Test()
{
MyData = new Data();
MyData.One = 1;
return View("Test",MyData)
}
In my TestView
<% Html.RenderPartial("PartialView",Model); %>
This give me a stackoverflow exception. So I am not sure how to pass it on. Of course I don't want to make the test view strongly typed if possible as what happens if I had like 10 strongly typed partial views in that view I would need like some sort of wrapper.
Is there a way to retrieve the client side ID for a model property similar to how the helpers like LabelFor etc generate the ID? For example, I have a model with a DateOfBirth property. I want to hook up that field to a jQuery DatePicker. Normally, I would hook up the date picker like this:
[Code]....
This works fine but hard codes the DateOfBirth property name. Ideally, I would like to use this:
[Code]....
Is this already available out of the box somewhere? I just created an extension method on the HtmlHelper class that does exactly this, but I wondered if there was already something built-in so I don't have to reinvent and test the wheel.
I found little bug feature in MVC 2.
So, if we have in route defined parameter with same name as property at model and then we call it from strongly typed HTML helper (for example Html.TexBoxFor(x => x.PropertyName)), we get value from route parameter instead of property of model.
For example.
We have defined follow route:
[Code]....
Very simple model:
[Code]....
Next action in controller:
[Code]....
And very simple view:
[Code]....
Result you can see at follow picture:
I think what this dirty example described the essence of the problem. Let me know if I'm wrong.
as the behaviour of the UpdateModel<T> changed in MVC 2 (RTM)? Since then I have the following problem when UpdateModel is called:UpdateModel<T1>(record, properties);record is typeof(Jumbo.Juist.Models.Bezorging) and properties is a string[] containing "Winkelnummer" and "OudRoutenummer".This now fails, because a get-property is being accessed: BerekendeAankomstTijd. In the past, it didn't fail. IMHO, this validation should be skipped, because it's only a getter (and thus/also not included in the string[] of properties to update)
View 14 RepliesI have a viewmodel with a property of a model class type. I set some properties in the model class to be able to show it to the user but then when I make a post, the model gets validated. How to overcome this? I don't want to enforce the DataAnnotations contraint in this case....
public class TheViewModel
{
TheModel TheModel { get; set;}
}
[code]...
Action:
[Code]....
View:
[Code]....
But error raised "object does not have property Name". Why?
I would like to exclude one of model's property from the Html.EditorFor. I tried [HiddenInput(DisplayValue=false)] but it renders the property as an hidden input. This could potentially be a security problem. How can we completely exclude a field from Html.EditorFor?
View 5 RepliesI got 3 radion-buttons on the view called:
- Taxi
- Train
- Bus
The model passed has a property called "Transport" of type String. In the action, if I set the property as 'Train', then the view should set the radio-button 'Train' checked. ie.
Action:
model.Transport = "Train";
How do I do that in the view to make the radio-button Train checked?
public class BandProfileModel
{
public BandModel Band { get; set; }
public IEnumerable<Relationship> Requests { get; set; }
}
and the following form:
<% using (Html.BeginForm()) { %>
<%: Html.EditorFor(m => m.Band) %>
<input type="submit" value="Save Band" />
<% } %>
which posts to the following action:
public ActionResult EditPost(BandProfileModel m, string band)
{
// stuff is done here, but m is null?
return View(m);
}
Basically, I only have one property on my model that is used in the form. The other property in BandProfleModel is just used in the UI for other data. I'm trying to update just the Band property, but for each post, the argument "m" is always null (specifically, the .Band property is null). It's posting just fine to the action, so it isn't a problem with my route. Just the data is null.
The ID and name attributes of the fields are BAND_whatever and Band.whatever (whatever being a property of Band), so it seems like it would work. What am I doing wrong? How can I use just one property as part of a form, post back, and have values populated via the model binder for my BandProfileModel property in the action?
I want to assign a stylesheet-class to a <div> element. The cssclass I want to use depends on the property 'Level' which is defined in de ViewModel.
I want to have something like this:
<div class="dashboardlevel" + <%= Model.Level %>>
But this doen't work.