I want to have 2 models in one view. The page contains both LoginViewModel and RegisterViewModel.
e.g.
public class LoginViewModel
{
public string Email { get; set; }
public string Password { get; set; }
}
public class RegisterViewModel
{
public string Name { get; set; }
public string Email { get; set; }
public string Password { get; set; }
}
Do I need to make another ViewModel which holds these 2 ViewModels?
public BigViewModel
{
public LoginViewModel LoginViewModel{get; set;}
public RegisterViewModel RegisterViewModel {get; set;}
}
I need the validation attributes to be brought forward to the view. This is why I need the ViewModels.
Isn't there another way such as (without the BigViewModel):
#model ViewModel.RegisterViewModel
#using (Html.BeginForm("Login", "Auth", FormMethod.Post))
{
#Html.TextBoxFor(model => model.Name)
#Html.TextBoxFor(model => model.Email)
#Html.PasswordFor(model => model.Password)
}
#model ViewModel.LoginViewModel
#using (Html.BeginForm("Login", "Auth", FormMethod.Post))
{
#Html.TextBoxFor(model => model.Email)
#Html.PasswordFor(model => model.Password)
}
There are lots of ways...
with your BigViewModel
you do:
#model BigViewModel
#using(Html.BeginForm()) {
#Html.EditorFor(o => o.LoginViewModel.Email)
...
}
you can create 2 additional views
Login.cshtml
#model ViewModel.LoginViewModel
#using (Html.BeginForm("Login", "Auth", FormMethod.Post))
{
#Html.TextBoxFor(model => model.Email)
#Html.PasswordFor(model => model.Password)
}
and register.cshtml same thing
after creation you have to render them in the main view and pass them the viewmodel/viewdata
so it could be like this:
#{Html.RenderPartial("login", ViewBag.Login);}
#{Html.RenderPartial("register", ViewBag.Register);}
or
#{Html.RenderPartial("login", Model.LoginViewModel)}
#{Html.RenderPartial("register", Model.RegisterViewModel)}
using ajax parts of your web-site become more independent
iframes, but probably this is not the case
I'd recommend using Html.RenderAction and PartialViewResults to accomplish this; it will allow you to display the same data, but each partial view would still have a single view model and removes the need for a BigViewModel
So your view contain something like the following:
#Html.RenderAction("Login")
#Html.RenderAction("Register")
Where Login & Register are both actions in your controller defined like the following:
public PartialViewResult Login( )
{
return PartialView( "Login", new LoginViewModel() );
}
public PartialViewResult Register( )
{
return PartialView( "Register", new RegisterViewModel() );
}
The Login & Register would then be user controls residing in either the current View folder, or in the Shared folder and would like something like this:
/Views/Shared/Login.cshtml: (or /Views/MyView/Login.cshtml)
#model LoginViewModel
#using (Html.BeginForm("Login", "Auth", FormMethod.Post))
{
#Html.TextBoxFor(model => model.Email)
#Html.PasswordFor(model => model.Password)
}
/Views/Shared/Register.cshtml: (or /Views/MyView/Register.cshtml)
#model ViewModel.RegisterViewModel
#using (Html.BeginForm("Login", "Auth", FormMethod.Post))
{
#Html.TextBoxFor(model => model.Name)
#Html.TextBoxFor(model => model.Email)
#Html.PasswordFor(model => model.Password)
}
And there you have a single controller action, view and view file for each action with each totally distinct and not reliant upon one another for anything.
Another way is to use:
#model Tuple<LoginViewModel,RegisterViewModel>
I have explained how to use this method both in the view and controller for another example: Two models in one view in ASP MVC 3
In your case you could implement it using the following code:
In the view:
#using YourProjectNamespace.Models;
#model Tuple<LoginViewModel,RegisterViewModel>
#using (Html.BeginForm("Login1", "Auth", FormMethod.Post))
{
#Html.TextBoxFor(tuple => tuple.Item2.Name, new {#Name="Name"})
#Html.TextBoxFor(tuple => tuple.Item2.Email, new {#Name="Email"})
#Html.PasswordFor(tuple => tuple.Item2.Password, new {#Name="Password"})
}
#using (Html.BeginForm("Login2", "Auth", FormMethod.Post))
{
#Html.TextBoxFor(tuple => tuple.Item1.Email, new {#Name="Email"})
#Html.PasswordFor(tuple => tuple.Item1.Password, new {#Name="Password"})
}
Note that I have manually changed the Name attributes for each property when building the form. This needs to be done, otherwise it wouldn't get properly mapped to the method's parameter of type model when values are sent to the associated method for processing. I would suggest using separate methods to process these forms separately, for this example I used Login1 and Login2 methods. Login1 method requires to have a parameter of type RegisterViewModel and Login2 requires a parameter of type LoginViewModel.
if an actionlink is required you can use:
#Html.ActionLink("Edit", "Edit", new { id=Model.Item1.Id })
in the controller's method for the view, a variable of type Tuple needs to be created and then passed to the view.
Example:
public ActionResult Details()
{
var tuple = new Tuple<LoginViewModel, RegisterViewModel>(new LoginViewModel(),new RegisterViewModel());
return View(tuple);
}
or you can fill the two instances of LoginViewModel and RegisterViewModel with values and then pass it to the view.
Use a view model that contains multiple view models:
namespace MyProject.Web.ViewModels
{
public class UserViewModel
{
public UserDto User { get; set; }
public ProductDto Product { get; set; }
public AddressDto Address { get; set; }
}
}
In your view:
#model MyProject.Web.ViewModels.UserViewModel
#Html.LabelFor(model => model.User.UserName)
#Html.LabelFor(model => model.Product.ProductName)
#Html.LabelFor(model => model.Address.StreetName)
Do I need to make another view which holds these 2 views?
Answer:No
Isn't there another way such as (without the BigViewModel):
Yes, you can use Tuple (brings magic in view having multiple model).
Code:
#model Tuple<LoginViewModel, RegisterViewModel>
#using (Html.BeginForm("Login", "Auth", FormMethod.Post))
{
#Html.TextBoxFor(tuple=> tuple.Item.Name)
#Html.TextBoxFor(tuple=> tuple.Item.Email)
#Html.PasswordFor(tuple=> tuple.Item.Password)
}
#using (Html.BeginForm("Login", "Auth", FormMethod.Post))
{
#Html.TextBoxFor(tuple=> tuple.Item1.Email)
#Html.PasswordFor(tuple=> tuple.Item1.Password)
}
Add this ModelCollection.cs to your Models
using System;
using System.Collections.Generic;
namespace ModelContainer
{
public class ModelCollection
{
private Dictionary<Type, object> models = new Dictionary<Type, object>();
public void AddModel<T>(T t)
{
models.Add(t.GetType(), t);
}
public T GetModel<T>()
{
return (T)models[typeof(T)];
}
}
}
Controller:
public class SampleController : Controller
{
public ActionResult Index()
{
var model1 = new Model1();
var model2 = new Model2();
var model3 = new Model3();
// Do something
var modelCollection = new ModelCollection();
modelCollection.AddModel(model1);
modelCollection.AddModel(model2);
modelCollection.AddModel(model3);
return View(modelCollection);
}
}
The View:
enter code here
#using Models
#model ModelCollection
#{
ViewBag.Title = "Model1: " + ((Model.GetModel<Model1>()).Name);
}
<h2>Model2: #((Model.GetModel<Model2>()).Number</h2>
#((Model.GetModel<Model3>()).SomeProperty
a simple way to do that
we can call all model first
#using project.Models
then send your model with viewbag
// for list
ViewBag.Name = db.YourModel.ToList();
// for one
ViewBag.Name = db.YourModel.Find(id);
and in view
// for list
List<YourModel> Name = (List<YourModel>)ViewBag.Name ;
//for one
YourModel Name = (YourModel)ViewBag.Name ;
then easily use this like Model
My advice is to make a big view model:
public BigViewModel
{
public LoginViewModel LoginViewModel{get; set;}
public RegisterViewModel RegisterViewModel {get; set;}
}
In your Index.cshtml, if for example you have 2 partials:
#addTagHelper *,Microsoft.AspNetCore.Mvc.TagHelpers
#model .BigViewModel
#await Html.PartialAsync("_LoginViewPartial", Model.LoginViewModel)
#await Html.PartialAsync("_RegisterViewPartial ", Model.RegisterViewModel )
and in controller:
model=new BigViewModel();
model.LoginViewModel=new LoginViewModel();
model.RegisterViewModel=new RegisterViewModel();
I want to say that my solution was like the answer provided on this stackoverflow page: ASP.NET MVC 4, multiple models in one view?
However, in my case, the linq query they used in their Controller did not work for me.
This is said query:
var viewModels =
(from e in db.Engineers
select new MyViewModel
{
Engineer = e,
Elements = e.Elements,
})
.ToList();
Consequently, "in your view just specify that you're using a collection of view models" did not work for me either.
However, a slight variation on that solution did work for me. Here is my solution in case this helps anyone.
Here is my view model in which I know I will have just one team but that team may have multiple boards (and I have a ViewModels folder within my Models folder btw, hence the namespace):
namespace TaskBoard.Models.ViewModels
{
public class TeamBoards
{
public Team Team { get; set; }
public List<Board> Boards { get; set; }
}
}
Now this is my controller. This is the most significant difference from the solution in the link referenced above. I build out the ViewModel to send to the view differently.
public ActionResult Details(int? id)
{
if (id == null)
{
return new HttpStatusCodeResult(HttpStatusCode.BadRequest);
}
TeamBoards teamBoards = new TeamBoards();
teamBoards.Boards = (from b in db.Boards
where b.TeamId == id
select b).ToList();
teamBoards.Team = (from t in db.Teams
where t.TeamId == id
select t).FirstOrDefault();
if (teamBoards == null)
{
return HttpNotFound();
}
return View(teamBoards);
}
Then in my view I do not specify it as a list. I just do "#model TaskBoard.Models.ViewModels.TeamBoards" Then I only need a for each when I iterate over the Team's boards. Here is my view:
#model TaskBoard.Models.ViewModels.TeamBoards
#{
ViewBag.Title = "Details";
}
<h2>Details</h2>
<div>
<h4>Team</h4>
<hr />
#Html.ActionLink("Create New Board", "Create", "Board", new { TeamId = #Model.Team.TeamId}, null)
<dl class="dl-horizontal">
<dt>
#Html.DisplayNameFor(model => Model.Team.Name)
</dt>
<dd>
#Html.DisplayFor(model => Model.Team.Name)
<ul>
#foreach(var board in Model.Boards)
{
<li>#Html.DisplayFor(model => board.BoardName)</li>
}
</ul>
</dd>
</dl>
</div>
<p>
#Html.ActionLink("Edit", "Edit", new { id = Model.Team.TeamId }) |
#Html.ActionLink("Back to List", "Index")
</p>
I am fairly new to ASP.NET MVC so it took me a little while to figure this out. So, I hope this post helps someone figure it out for their project in a shorter timeframe. :-)
Create one new class in your model and properties of LoginViewModel and RegisterViewModel:
public class UserDefinedModel()
{
property a1 as LoginViewModel
property a2 as RegisterViewModel
}
Then use UserDefinedModel in your view.
you can always pass the second object in a ViewBag or View Data.
This is a simplified example with IEnumerable.
I was using two models on the view: a form with search criteria (SearchParams model), and a grid for results, and I struggled with how to add the IEnumerable model and the other model on the same view. Here is what I came up with, hope this helps someone:
#using DelegatePortal.ViewModels;
#model SearchViewModel
#using (Html.BeginForm("Search", "Delegate", FormMethod.Post))
{
Employee First Name
#Html.EditorFor(model => model.SearchParams.FirstName,
new { htmlAttributes = new { #class = "form-control form-control-sm " } })
<input type="submit" id="getResults" value="SEARCH" class="btn btn-primary btn-lg btn-block" />
}
<br />
#(Html
.Grid(Model.Delegates)
.Build(columns =>
{
columns.Add(model => model.Id).Titled("Id").Css("collapse");
columns.Add(model => model.LastName).Titled("Last Name");
columns.Add(model => model.FirstName).Titled("First Name");
})
...
)
SearchViewModel.cs:
namespace DelegatePortal.ViewModels
{
public class SearchViewModel
{
public IEnumerable<DelegatePortal.Models.DelegateView> Delegates { get; set; }
public SearchParamsViewModel SearchParams { get; set; }
....
DelegateController.cs:
// GET: /Delegate/Search
public ActionResult Search(String firstName)
{
SearchViewModel model = new SearchViewModel();
model.Delegates = db.Set<DelegateView>();
return View(model);
}
// POST: /Delegate/Search
[HttpPost]
public ActionResult Search(SearchParamsViewModel searchParams)
{
String firstName = searchParams.FirstName;
SearchViewModel model = new SearchViewModel();
if (firstName != null)
model.Delegates = db.Set<DelegateView>().Where(x => x.FirstName == firstName);
return View(model);
}
SearchParamsViewModel.cs:
namespace DelegatePortal.ViewModels
{
public class SearchParamsViewModel
{
public string FirstName { get; set; }
}
}
I have this ViewModel
public class MyViewModel
{
public Customer Customer{ get; set; }
public Account Account{ get; set; }
public DateTime MyDate{ get; set; }
}
This View
#using (Html.BeginForm("Final", "Home", FormMethod.Post, new { #class = "form" }))
{
#Html.AntiForgeryToken()
#Html.HiddenFor(m => m.Customer)....
#Html.DisplayFor(m => m.Customer.FirstName) //This displays the name ok
And this is my Controller
[HttpPost]
[ValidateAntiForgeryToken]
public ActionResult Final(MyViewModel viewModel)
{
viewModel.Customer.... //This is null
The problem is, I don´t have to edit the properties, only show them and then save to the database, that´s why I use HiddenFor, but the properties Customer and Account bind to null.
What is the problem? Maybe the HiddenFor?
PS: I have the GET method in which I perform the appropriate return View(viewModel)
This is where I pass the ViewModel
[HttpGet]
public ActionResult Final()
{
var viewModel = new MyViewModel
{
//set the properties, etc
};
return View(viewModel);
}
Customer is a complex object, so you will need each property of Customer as a hidden field.
#Html.HiddenFor(m => m.Customer.FirstName)
#Html.HiddenFor(m => m.Customer.LastName)
Alternatively, you can store the entire MyViewModel in TempData which uses Session State under the hood.
When I have the following code and I submit the form my post action shows the Contact object as null. If I remove the DropDownListFor from the view the Contact object contains the expected information (FirstName). Why? How do I get the SelectList value to work?
My classes:
public class ContactManager
{
public Contact Contact { get; set; }
public SelectList SalutationList { get; set; }
}
public class Contact
{
public int Id{get;set;}
public string FirstName{get; set;}
public SalutationType SalutationType{get; set;}
}
public class SalutationType
{
public int Id { get; set; }
public string Name { get; set; }
}
My view:
#model ViewModels.ContactManager
#using (Html.BeginForm())
{
#Html.AntiForgeryToken()
#Html.ValidationSummary(true)
#Html.HiddenFor(model => model.Contact.Id)
#Html.DropDownListFor(model => model.Contact.SalutationType.Id, Model.SalutationList, "----", new { #class = "form-control" })
#Html.EditorFor(model => model.Contact.FirstName)
<input type="submit" value="Save" />
}
My Controller:
public ActionResult Edit(int? id)
{
Contact contact = db.Contacts.FirstOrDefault(x => x.Id == id);
ContactManager cm = new ContactManager();
cm.Contact = contact;
cm.SalutationList = new SelectList(db.SalutationTypes.Where(a => a.Active == true).ToList(), "Id", "Name");
return View(cm);
}
[HttpPost]
public ActionResult Edit(ContactManger cm)
{
//cm at this point is null
var test = cm.Contact.FirstName;
return View();
}
You will pass the DropdownList using ViewBag:
ViewBag.SalutationList = new SelectList(db.SalutationTypes.Where(a => a.Active == true).ToList(), "Id", "Name");
than u have to call this list inside your edit view:
#Html.DropDownList("SalutationList",String.Empty)
The problem is that the DefaultModelBinder won't be able to map nested models properly if you use a different parameter name. You must use the same parameter name as the model name.
public ActionResult Edit(ContactManager contactManager)
As a general practice, always use the name of the model as the parameter name to avoid mapping problems.
Further Suggestion:
You can just use Contact as the parameter model, no need to use ContactManager if you only need the contact model.
[HttpPost]
public ActionResult Edit(Contact contact)
{
var test = contact.FirstName;
return View();
}
I have a Register Primary View which shows two different types of Addresses 1. Home Address 2. Mailing Address
public class RegisterModel
{
public AddressModel HomeAddress { get; set; }
public AddressModel MailAddress { get; set; }
}
public class AddressModel
{
public string Street1 { get; set; }
public string Street2 { get; set; }
public string State { get; set; }
public string City { get; set; }
}
My main Register View is Strongly Typed to RegisterModel as follows
#model MyNamespace.Models.RegisterModel
#{
Layout = "~/Views/_Layout.cshtml";
}
#using (Html.BeginForm(null, null, FormMethod.Post, new { id = "myForm" }))
{
<div id="form">
#Html.Action("MyAddressPartial")
#Html.Action("MyAddressPartial")
</div>
}
MyAddressPartialView as follows : -
#model MyNamespace.Models.AddressModel
#{
Layout = "~/Views/_Layout.cshtml";
}
<div id="Address">
#Html.TextBoxFor(m=>m.Street1 ,new { #id="Street1 "})
#Html.TextBoxFor(m=>m.Street2,new { #id="Street2"})
#Html.TextBoxFor(m=>m.State ,new { #id="State "})
#Html.TextBoxFor(m=>m.City,new { #id="City"})
</div>
My RegisterController:-
// Have to instantiate the strongly Typed partial view when my form first loads
// and then pass it as parameter to "Register" post action method.
// As you can see the #Html.Action("MyAddressPartial") above in main
// Register View calls this.
public ActionResult MyAddressPartial()
{
return PartialView("MyAddressPartialView", new AddressModel());
}
I submit my Main Form to below mentioned action method in same Register Controller.
[HttpPost]
public ActionResult Register(RegisterModel model,
AddressModel homeAddress,
AddressModel mailingAddress)
{
//I want to access homeAddress and mailingAddress contents which should
//be different, but as if now it comes same.
}
I don't want to create a separate class one for MailingAddress and one for HomeAddress. if I do that then I will have to create two separate strongly typed partial views one for each address.
Any ideas on how to reuse the classes and partial views and make them dynamic and read their separate values in Action Method Post.
Edit 1 Reply to scott-pascoe:-
In DisplayTemplates Folder, I added following AddressModel.cshtml
<div>
#Html.DisplayFor(m => m.Street1);
#Html.DisplayFor(m => m.Street2);
#Html.DisplayFor(m => m.State);
#Html.DisplayFor(m => m.City);
</div>
Also In EditorTemplate Folder, I added following AddressModel.cshtml but with EditorFor
<div>
#Html.EditorFor(m => m.Street1);
#Html.EditorFor(m => m.Street2);
#Html.EditorFor(m => m.State);
#Html.EditorFor(m => m.City);
</div>
Now how do i use them in RegisterView and also how i read values in Controller's post Action Method ? What else would have to be modified ? I have added almost entire code above. I am pretty beginner to MVC.
The typical ASP.NET MVC method for doing this is to use EditorTemplates and DisplayTemplates for your custom types.
In ~/Views/Shared, Create two folders, DisplayTemplates, and EditorTemplates.
In the DisplayTemplates folder create a partial view with the name of your Model, ie (AddressModel), and create a DisplayFor Template.
In the EditorTemplates folder create another partial view named AddressModel.cshtml and create an EditorFor Template.
MVC will then automatically use your templates and give you the data that you are asking for.
Use #Html.EditorFor (or #Html.DisplayFor, for display) in your view:
#model MyNamespace.Models.RegisterModel
#{
Layout = "~/Views/_Layout.cshtml";
}
#using (Html.BeginForm(null, null, FormMethod.Post, new { id = "myForm" }))
{
<div id="form">
#Html.EditorFor(m => m.HomeAddress)
#Html.EditorFor(m => MailAddress)
</div>
}
You will not need to have a separate controller action for the parts, just populate the addresses in the RegisterModel before in your controller. Like this:
[HttpGet]
public ActionResult Register() // this will be the page people see first
{
var model = new RegisterModel();
return View(model); // assuming your view is called Register.cshtml
}
[HttpPost]
public ActionResult Register(RegisterModel model){
DosomethingWithHomeAddress(model.HomeAddress);
DosomethingWithMailAddress(model.MailAddress);
model.IsSaved = true; // some way to let the user knwo that save was successful;
// if this is true, display a paragraph on the view
return View(model);
}
I have a Model Wrapper class that wrap another model:
public class LogicPage {
public MyPage Pg { get; set; }
}
public class MyPage {
public string name { get; set; }
}
In my create form view I use strongly typed form helper:
#model MyAppCore.Components.LogicPage
#using (Html.BeginForm("Create","PageAdmin",FormMethod.Post)) {
<div class="editor-field">
#Html.EditorFor(model => model.Pg.name)
#Html.ValidationMessageFor(model => model.Pg.name)
</div>
...
}
In my controller, I define the action as:
[HttpPost]
public ActionResult Create(LogicPage logicpg)
{
return View("View",logicpg.Pg);
}
However, when the MyPage Pg is displayed in the "View" which list details of the model (MyPage), no user inputted data are shown, the details of the page shown are all MyPage default values (initialized with default no parameter constructor of LogicPage). Looks like data in the form are not passing to the model in the action. Can someone please help me why the data is not passing?
To clarify more, I have two views "Create" form view of model LogicPage and the "View" detail view for model MyPage
Create.cshtml
#model MyAppCore.Components.LogicPage
View.cshtml
#model MyAppCore.Components.MyPage
Thanks
You say you're doing
[HttpPost]
public ActionResult Create(LogicPage logicpg)
{
return View("View",logicpg.Pg);
}
But given that MyPage doesn't inherit from LogicPage, and your view expects an instance of LogicPage, I think you should be doing:
[HttpPost]
public ActionResult Create(LogicPage logicpg)
{
return View("View",logicpg);
}