How can I pass all my data to PartialView.
My table code look like this.
#foreach (var item in Model.listContract)
{
<tr style="font-size:11px;">
<td>View more info</td>
<td>#item.ID</td>
<td>#item.name</td>
<td>#item.Phone</td>
<td>#item.Passport</td>
<td>#item.CustomerID</td>
</tr>
}
From Controller bind data to ViewModel pass to View
public ActionResult MethodName(){
var viewmodel = new ViewModel();
//bind data to ViewModel
return View(viewmodel)
}
In View pass partial view data as list model
View :
#model ViewModel
#Html.Partial("PartialViewName", Model.DataList)
Partial View :
#model List<Model>
#foreach(var item in Model){
.....
}
ViewModel :
public class ViewModel {
//properties for View
....
//property for partial view
public List<Model> DataList {get; set;}
}
Related
I have this error on read my view on #foreach.
Exception Details: System.NullReferenceException: Object reference not set to an instance of an object. Model is null.
this is my userprofile model.
public class UsersContext : DbContext
{
public UsersContext()
: base("DefaultConnection")
{
}
public DbSet<UserProfile> UserProfiles { get; set; }
public DbSet<Secretarias> Secretarias { get; set; }
public DbSet<Secretarias_Direcciones> Secretarias_Direcciones { get; set; }
}
[Table("UserProfile")]
public class UserProfile
{
[Key]
public int UserId { get; set; }
[Required]
public string Nombre { get; set; }
[Required]
public int SecretariaId { get; set; }
[Required]
public int DireccionId { get; set; }
[Required]
public string UserName { get; set; }
}
this is my control
public ActionResult ListaUsuarios()
{
UsersContext db = new UsersContext();
var model = from usrs in db.UserProfiles
select new { usrs.UserId, usrs.Nombre, usrs.UserName, usrs.SecretariaId, usrs.DireccionId };
ViewBag.Model = model.ToList();
return View();
}
this is my view
#foreach (var item in Model)
{
<tr>
<td>
#Html.DisplayFor(modelItem => item.UserId)
</td>
<td>
#Html.DisplayFor(modelItem => item.Nombre)
</td>
<td>
#Html.DisplayFor(modelItem => item.UserName)
</td>
<td>
#Html.DisplayFor(modelItem => item.SecretariaId)
</td>
<td>
#Html.DisplayFor(modelItem => item.DireccionId)
</td>
<td>
#Html.ActionLink("Edit", "Edit", new { id=item.UserId }) |
#Html.ActionLink("Details", "Details", new { id=item.UserId}) |
#Html.ActionLink("Delete", "Delete", new { id=item.UserId })
</td>
</tr>
}
Here are the Exception Details:
System.NullReferenceException: Object reference not set to an instance
of an object. Model is null.
You're getting the NullReferenceException because you haven't passed a strongly-typed model to your view. So in your foreach loop, Model is null. The normal way of doing that is like so:
return View(model);
The real problem is that you're mixing different concepts here. You're trying to access the properties of a model in a strongly-typed way, and yet you're defining a ViewBag.Model property, which is dynamic and has no relation whatsoever to the Model property within a view.
If you want to access your model in a strongly-typed way, and that's certainly the way I'd suggest you do it, you need to make several changes. I'd suggest you first define a view model to represent the data for a UserProfile:
public class UserProfileViewModel
{
public int UserId { get; set; }
public string Nombre { get; set; }
public int SecretariaId { get; set; }
public int DireccionId { get; set; }
public string UserName { get; set; }
}
Now, change your action to create a list of these:
public ActionResult ListaUsuarios()
{
using (UsersContext db = new UsersContext())
{
var model = from usrs in db.UserProfiles
select new UserProfileViewModel
{
UserId = usrs.UserId,
Nombre = usrs.Nombre,
UserName = usrs.UserName,
SecretariaId = usrs.SecretariaId,
DireccionId = usrs.DireccionId
};
return View(model.ToList());
}
}
Notice that I've made several changes here. Firstly, I've added the using statement to ensure UsersContext is disposed properly. Next, I've changed the query to instantiate UserProfileViewModels, instead of anonymous types. Lastly, I'm directly passing model as a parameter to View(), making sure to call ToList() on it, in order to evaluate the query before UsersContext gets disposed.
Finally, you just need to make one change to your view:
#model List<UserProfileViewModel>
#foreach (var item in Model)
{
// rest of code goes here
}
The #model directive specifies the exact model type that the view should expect to receive, thus making it strongly-typed.
Obviously - check the data types of each of the Ids that you are using to perform your joins in your database. Also ensure the database data types were mapped to the correct .Net data types.
If the data types of your ID fields appear to be ok, I would suggest narrowing down which join is causing the error. Modify you LINQ expression to have a single join and ensure there is no error. Add the other joins in one at a time until the error appears.
For your model, please check that it is being populated.
Depending on whatever is happening, it could be that your model is just null
and possibly its type might be something that it is not expected.
-- in the controller a foreach loop on the model context reference will let you see the values
I have a model:
public class TestModel {
[Display(Name = "Date")]
public DateTime Date { get; set; }
}
with Html.LabelFor helper method in Test.cshtml page
#Html.LabelFor(m => m.Date )
and use this page with 2 MVC action methods: Create and Update.
Example:
public virtual ViewResult Create() {
return View("Test");
}
public virtual ViewResult Update() {
return View("Test");
}
and I want to display #Html.LabelFor(m => m.Date ) with Create page: "Date" and Update page: "Update Date" . I know if the normal way of MVC3 can't do this. I hope your ideal can edit Html.LabelFor hepler method or anything way to bind data to Html.LabelFor in action methods on the controller
Adding a hiddenFor field will bind the data to your Model.
#Html.HiddenFor(m=>m.Date);
For override, please just look this answer
https://stackoverflow.com/a/5196392/5557777
you can override editorfor like this How can I override the #Html.LabelFor template? but I think you can do it more easily with ViewBag:
public virtual ViewResult Create() {
ViewBag.Title = "Create";
return View("Test");
}
public virtual ViewResult Update() {
ViewBag.Title = "Update";
return View("Test");
}
in view:
#string.format("{0} Date" , ViewBag.Title )
I have this code in mvc 3 razor
#using (Html.BeginForm("MyAction", "MyController"))
{
<input type="text" id="txt" name="txt"/>
<input type="image" src="image.gif" alt="" />
}
in controller I have this code
[HttpPost]
public ActionResult MyAction(string text)
{
//TODO something with text and return value...
}
now, how to send a new value, for exemple id to Action result??? Thanks
You use a view model:
public class MyViewModel
{
public string Text { get; set; }
// some other properties that you want to work with in your view ...
}
and then pass this view model to the view:
public ActionResult MyAction()
{
var model = new MyViewModel();
model.Text = "foo bar";
return View(model);
}
[HttpPost]
public ActionResult MyAction(MyViewModel model)
{
// remove the Text property from the ModelState if you intend
// to modify it in the POST controller action or HTML helpers will
// use the old value
ModelState.Remove("Text");
model.Text = "some new value";
return View(model);
}
and then the view is strongly typed to this model:
#model MyViewModel
#using (Html.BeginForm("MyAction", "MyController"))
{
#Html.EditorFor(x => x.Text)
<input type="image" src="image.gif" alt="" />
}
I try to initialize the DataView.Model in a partial view. The Page works fine but when I return to the controller the model is empty.
some help(solution or an explanation why it is not right).
thanks!!
code:
In my Partial View:
ViewData.Model = new DiamondPrint();
ViewData.Model.Diamond = m_db.DiamondInfoes.Where(di => di.Id == id).SingleOrDefault();
In my Controller:
public ActionResult Preview(DiamondPrint d)//the properties in d = null
{
return View(d);
}
Here is a great article on Model Binding. Model Binding Make sure you are setting the name property in your html input fields.
Looking at the code you have included it seems that you are initialising the ViewData.Model in the partial view but in the controller action you are expecting the default model binder to recreate your model. For the model binder to recreate your model you will need to have created a strongly typed view.
For example:
Controller:
public ActionResult Index()
{
return View();
}
[HttpPost]
public ActionResult Index(IndexModel model)
{
return View();
}
Model:
public class IndexModel
{
public string MyValue { get; set; }
}
View:
Note the #model definition at the top (ignore namespace)
#model MvcApplication14.Models.IndexModel
#using (Html.BeginForm())
{
#Html.Partial("_IndexPartial", Model)
<input type="submit" value="click"/>
}
Partial View:
#model MvcApplication14.Models.IndexModel
#Html.EditorFor(m => m.MyValue)
I'm currently experiencing a weird issue with ASP.NET MVC 3 ListBox validation, as stated in the title. Basically, I have a List in my viewmodel, which I bind to a ListBox with multiple selection enabled.
The List is given an attribute [Required]. When I submit the form with single value selected, it passes validation with no hiccups. However, with more than one, validation would fail.
Any thoughts?
Weird, I am unable to reproduce your issue.
Model:
public class MyViewModel
{
[Required(ErrorMessage = "Please select at least one item")]
public string[] SelectedItems { get; set; }
public IEnumerable<SelectListItem> Items
{
get
{
return Enumerable.Range(1, 5).Select(x => new SelectListItem
{
Value = x.ToString(),
Text = "item " + x
});
}
}
}
Controller:
public class HomeController : Controller
{
public ActionResult Index()
{
return View(new MyViewModel());
}
[HttpPost]
public ActionResult Index(MyViewModel model)
{
return View(model);
}
}
View:
#model MyViewModel
#using (Html.BeginForm())
{
#Html.ListBoxFor(x => x.SelectedItems, Model.Items)
#Html.ValidationMessageFor(x => x.SelectedItems)
<button type="submit">OK</button>
}
If you don't select any item in the list the validation error message is shown as expected. If you select one or more items the validation passes and no error message is displayed.