I am trying to get a HTTPPost working for a comment section of a video page but am having some difficulty on postback. The save is being recorded to the database but on reload of the page I am not able to find the @Model.SelectedMediaItem.Name variable that was found the first time. My html is as follows:
@{
ViewBag.Title = "Screencast";
Layout = "~/Views/Shared/_Layout.cshtml";
}
@model Project.ViewModels.MediaViewModel
<body onload="initialize()">
<div class="container region3wrap_screencast">
<div class="row content_top_contact">
<div class="nine columns">
<ul class="breadcrumbs">
<li><a href="@Url.Action("Index","Home")">Home</a></li>
<li><a href="@Url.Action("Index","Media")">Media</a></li>
<li><a href="@Url.Action("Screencast","Media")">Screencast</a></li>
<li class="current"><a href="#">@Model.SelectedMediaItem.Name</a></li>
</ul>
</div>
</div>
</div>
<div class="twelve columns leave-comment">
<h3>Leave a Comment</h3>
@using(Html.BeginForm("Screencast","Media", FormMethod.Post, Model))
{
<div class="row">
<div class="six columns">
<div class="row">
<div class="six columns">
@Html.LabelFor(c => c.FeedbackComment.UserID)
@Html.TextBoxFor(c => c.FeedbackComment.UserID)
</div>
<div class="six columns">
@Html.LabelFor(c => c.FeedbackComment.ContentID)
@Html.TextBoxFor(c => c.FeedbackComment.ContentID)
</div>
<div class="row">
<div class="twelve columns">
@Html.LabelFor(c => c.FeedbackComment.FeedbackString)
@Html.TextAreaFor(c => c.FeedbackComment.FeedbackString)
</div>
</div>
</div>
</div>
</div>
<input type="submit" value="Submit button" class="medium button bottom20"/>
}
</div>
</body>
My controller actions are as follows:
//GET
public ActionResult Screencast(int ID)
{
mvm = new ViewModels.MediaViewModel(ID);
return View(mvm);
}
//POST
[HttpPost]
public ActionResult Screencast(MediaViewModel mvm)
{
Feedback newFeedback= new Feedback();
newFeedback.UserID = mvm.FeedbackComment.UserID;
newFeedback.ContentID = mvm.FeedbackComment.ContentID;
newFeedback.CourseID = null;
newFeedback.Timestamp = DateTime.Now;
newFeedback.FeedbackString = mvm.FeedbackComment.FeedbackString;
//Initialize the Feedback Repository and save changes
feedbackRepository = new Repository<Feedback>(dbcontext);
feedbackRepository.Add(newFeedback);
feedbackRepository.SaveChanges();
return View(mvm);
}
When I call the page to begin with at the URL /Media/Screencast/1 it loads with a viewmodel populated with details of the SelectedMediaItem and all appears as is expected.
When I try to post back to this my viewmodel detail is being lost although the comment is actually saved as expected I need to visit another page and revisit before I am able to see.
How can I pass my view model in the @Html.BeginFor as an additional parameter?
My ViewModel is as follows:
public class MediaViewModel
{
private Repository<Content> contentRepository;
private Repository<Feedback> feedbackRepository;
private MetaLearningContext dbcontext;
public int screencastID { get; set; }
public IEnumerable<Content> Screencasts { get; set; }
public IEnumerable<Content> Podcasts { get; set; }
public IEnumerable<Content> Documents { get; set; }
public IEnumerable<Feedback> FeedbackComments { get; set; }
public Content SelectedMediaItem { get; set; }
public MetaLearningUser User { get; set; }
public MetaLearningUser FeedbackAuthor { get; set; }
public Feedback FeedbackComment { get; set; }
public MediaViewModel()
{
this.dbcontext = new MyContext(System.Configuration.ConfigurationManager.ConnectionStrings["MyContext"].ConnectionString);
this.contentRepository = new Repository<Content>(dbcontext);
this.feedbackRepository = new Repository<Feedback>(dbcontext);
this.dbcontext.Configuration.LazyLoadingEnabled = true;
//Retrieve a list of Screencasts
Screencasts = contentRepository
.Get(c => c.ContentTypeID == 1);
//Retrieve a list of Podcasts
Podcasts = contentRepository
.Get(c => c.ContentTypeID == 2);
//Retrieve a list of Documents
Documents = contentRepository
.Get(c => c.ContentTypeID == 3);
}
public MediaViewModel(int id)
{
this.dbcontext = new MyContext(System.Configuration.ConfigurationManager.ConnectionStrings["MyContext"].ConnectionString);
this.contentRepository = new Repository<Content>(dbcontext);
this.feedbackRepository = new Repository<Feedback>(dbcontext);
this.dbcontext.Configuration.LazyLoadingEnabled = true;
//Retrieve a list of Screencasts
Screencasts = contentRepository
.Get(c => c.ContentTypeID == 1);
//Retrieve a list of Podcasts
Podcasts = contentRepository
.Get(c => c.ContentTypeID == 2);
//Retrieve a list of Documents
Documents = contentRepository
.Get(c => c.ContentTypeID == 3);
//Retrieve selected screencast
SelectedMediaItem = contentRepository
.Get(c => c.ContentID == id)
.FirstOrDefault();
}
}
Just to add something to the answer from Mattias:
He's right - if you want some information where to redirect place it in a hidden field or add it as parameter to the Form-Method.
The pattern, which is common in MVC is PRG (Post/Redirect/Get) so essentially your POST should answer with a RedirectToAction("Screencast") instead of view. (http://en.wikipedia.org/wiki/Post/Redirect/Get)
To use this you should refactor your functions a little bit:
public ActionResult Screencast(int ID)
{
mvm = new ViewModels.MediaViewModel(ID); //The id is in the viewmodel i think
return View(mvm);
}
[HttpPost]
//Even if unused include the formcollection for cases where the GET has the same signature
public ActionResult Screencast(int ID, MediaViewModel mvm, FormCollection collection)
{
//Do something
return RedirectToAction("Screencast", new { id = ID });
}
And inside your view:
@using(Html.BeginForm("Screencast", new { id = Model.SelectedMediaItem }))
{ ... }