Question

i’ve a table

 post(p_id,u_id(FK),date,title,type,viewer)

and a Modelclass UserModel.cs with(p_id,u_id,date,title,type). When i to create a new post in a post controller i tried like

 public ActionResult Create(PostModel p)
        { 


        if (ModelState.IsValid)
        {
          p.date =  DateTime.Today;
          post post = new Models.DB.post();
          post.u_id = User.Identity.Name;
          post.date = p.date;
          post.title = p.title;
          post.type= p.type;
          post.viwer=1;

          try{      db.posts.AddObject(post);
                db.SaveChanges();
         return RedirectToAction("LoggedIndex");
         }
        catch{ return RedirectToAction("Index"); }
      }
    return View(p);
   }


It is not Working. I think from is not submitting when i submit always view is returned. i  i’ve created  stronly typed  create view. in create.cshtml i’ve two hidden field user_id and date of postModel Object. I’m very new at MVC.please help me thanks in advance….
												

Answer 1

Sending Data from View to the Controler by using the MVC-Model

I understand it like that: you have a mvc3 page and want to send some custom data to your controller:

Controller:

[http-post]
    public ActionResult Create(PostModel p)
        { 
        if (ModelState.IsValid)
        {
          p.date =  DateTime.Today;
          post post = new Models.DB.post();
          post.u_id = User.Identity.Name;
          post.date = p.date;
          post.title = p.title;
          post.type= p.type;
          post.viwer=1;

                db.posts.AddObject(post);
                db.SaveChanges();
         return RedirectToAction("LoggedIndex");
      }
    return View(p);
   }

public ActionResult Create()
        { 

    return View(new post());
   }

View:

@model post

@using (Html.BeginForm("Create", "Controller-Name", FormMethod.Post, new { id = "FORM" }))
{
 @Html.EditorFor(model => model.date)

 @Html.EditorFor(model => model.name)

 @Html.EditorFor(model => model.title)

 @Html.EditorFor(model => model.type)
}

Further informations: http://weblogs.asp.net/scottgu/archive/2010/07/27/introducing-asp-net-mvc-3-preview-1.aspx