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 a view,model and controller in my application. What I need is to pass the textbox value to a view and display on the result page.
Index.cshtml
#model MvcApplication5.Models.Employee
#{
ViewBag.Title = "Dashboard";
}
<script src="https://code.jquery.com/jquery-1.10.2.js"></script>
<h2>Dashboard</h2>
<div>
<h1>Employee </h1>
</div>
#using (Html.BeginForm("DisplayForm", "Dashboard", FormMethod.Post))
{
#Html.EditorFor(model =>model.username)
#Html.TextBoxFor(model => model.Company)
<button type="submit" id="ajax_method">submit </button>
}
Controller
public ActionResult DisplayForm(Employee model)
{
var employeeName = model.username;
var company = model.Company;
return View("Validateresult");
}
Model
public class Employee
{
public string username { get; set; }
public string Company { get; set; }
}
and validateresult.cshtml
#model MvcApplication5.Models.Employee
<h2>Validateresult</h2>
<div>
<div>username:
#Html.DisplayFor(model => model.username)
</div>
</div>
It seems a silly issue but im unable to find the reason why it's not displaying the username that i enter in to the textbox. The value is accessible in the controller but not in the view.
You're not passing the model to the view.
In controller, change return View("Validateresult"); to return View("Validateresult", model);
So, im currently building an application that needs the user model validating, and if the incorrect properties are filled in to the user it will tell them.
I have the data annotations set up, but im not sure how i relay the error message back to the user?
I have this set up so far on my model and view.
Model
public class DatabaseModel
{
[Required(ErrorMessage = ("A first name is required"))]
public string FirstName { get; set; }
[Required(ErrorMessage = ("A last name is required"))]
public string LastName { get; set; }
[Required(ErrorMessage = ("A valid role is required"))]
public string Role { get; set; }
// TODO - Validate rank to only b 1 - 10
//
[Range(1,10, ErrorMessage = ("A rank between 1 and 10 is required"))]
public int Rank { get; set; }
}
And View
#model RoleCreatorAndEditor.Models.DatabaseModel
#{
ViewData["Title"] = "Index";
}
<h2>User Information</h2>
<p>This is your user information!</p>
#using (Html.BeginForm("Index", "Home", FormMethod.Post)) {
#Html.Label("First Name")
<br>
#Html.TextBoxFor(m => m.FirstName)
<br>
#Html.Label("Last Name")
<br>
#Html.TextBoxFor(m=>m.LastName)
<br>
#Html.Label("Role")
<br>
#Html.TextBoxFor(m => m.Role)
<br>
#Html.Label("Rank")
<br>
#Html.TextBoxFor(m => m.Rank)
<br><br>
<input type="submit" value="Save">
}
My Controller
public class HomeController : Controller
{
// GET: Home
[HttpGet]
public ActionResult Index()
{
DatabaseModel model = new DatabaseModel();
return View(model);
}
[HttpPost]
public ActionResult Index(DatabaseModel model)
{
if (ModelState.IsValid)
{
ListToDatatable convert = new ListToDatatable();
DataTable user = convert.Convert(model);
DatabaseRepository dbRepo = new DatabaseRepository();
dbRepo.Upload(user);
}
return View();
}
}
I believe the model needs to be passed back to the view in order to display the error message, and although i have read through the documentation on asp.net i cannot understand how they just add the error message and the form knows how to display the errors to the user.
I am extremely confused.
You need to use ModelState.IsValid in your Controller and also #Html.ValidationMessageFor(model => model.FirstName) in your view:
public ActionResult Index(ViewModel _Model)
{
// Checking whether the Form posted is valid one.
if(ModelState.IsValid)
{
// your model is valid here.
// perform any actions you need to, like database actions,
// and/or redirecting to other controllers and actions.
}
else
{
// redirect to same action
return View(_Model);
}
}
For your example:
#model RoleCreatorAndEditor.Models.DatabaseModel
#{
ViewData["Title"] = "Index";
}
<h2>User Information</h2>
<p>This is your user information!</p>
#using (Html.BeginForm("Index", "Home", FormMethod.Post)) {
#Html.LabelFor(m=>m.FirstName)
<br>
#Html.TextBoxFor(m => m.FirstName)
#Html.ValidationMessageFor(model => model.FirstName, "", new { #class = "text-danger" })
<br>
#Html.LabelFor(m=>m.LastName)
<br>
#Html.TextBoxFor(m=>m.LastName)
#Html.ValidationMessageFor(model => model.LastName, "", new { #class = "text-danger" })
. . .
<input type="submit" value="Save">
}
Controller:
[HttpPost]
public ActionResult Index(DatabaseModel model)
{
if (ModelState.IsValid)
{
ListToDatatable convert = new ListToDatatable();
DataTable user = convert.Convert(model);
DatabaseRepository dbRepo = new DatabaseRepository();
dbRepo.Upload(user);
}
return View(model);
}
I am creating a web application where there are five steps.
Home Page1 Page 2 Review Confirmation.
In the url, it goes like localhost:22112/Home/Page1 Page 2 and so forth.
My problem is if someone copies localhost:22112/Home/Page2, then it skips everything
and jumps to page 2 directly. So, how can I stop that? I did the following but its not working properly.
Any suggestions would be really helpful.
In the controller
private bool IsFromIndexPage()
{
if (Session["IsFromIndex"] != null)
{
return true;
}
else
{
return false;
}
}
And for each page actionresult, I am writing it like this.
[HttpGet]
public ActionResult Page1()
{
if (!IsFromIndexPage())
{
return RedirectToAction("Index");
}
.....other methods..
}
[HttpPost]
public ActionResult Page1(Information model, string command)
{
if (!IsFromIndexPage())
{
return RedirectToAction("Index");
}
.....other methods..
}
[HttpGet]
public ActionResult Page2()
{
if (!IsFromIndexPage())
{
return RedirectToAction("Index");
}
.....other methods..
}
[HttpPost]
public ActionResult Page2(Information model, string command)
{
if (!IsFromIndexPage())
{
return RedirectToAction("Index");
}
.....other methods..
}
If you're using session to store the progress through the steps you should be checking your session variables to validate the request is for the given page otherwise redirect the user to the first/current completed page.
You can write a custom request handler for this to keep your session validation code separate your controller code
see this Question about how to implement the basic functionality to what you want to do
EDIT:
switch(currentStep){
case 1:
return Step1(model)
break;
case 2:
return Step2(model)
break;
default:
return new HttpNotFoundResult();
break;
}
Here is a bit different approach, on how to make a wizard with asp.net MVC using ajax.
Your url will be /Home/Wizard on every step. Since using the AjaxOnly attribute, it will not be possible to visit Step1, Step2 etc (see reference in the bottom for AjaxOnly)
Controller:
public ActionResult Wizard()
{
return View();
}
[AjaxOnly]
public ActionResult Step1()
{
return PartialView("Step1");
}
[AjaxOnly]
public PartialViewResult Step2(FormCollection coll)
{
Session["FullName"] = coll["FullName"]!= null ? coll["FullName"].ToString() : string.Empty;
return PartialView("Step2");
}
[AjaxOnly]
public PartialViewResult Confirm(FormCollection coll)
{
WizardModel model = new WizardModel() { Name = Session["FullName"].ToString(), Phone = coll["Phone"] != null ? coll["Phone"].ToString() : string.Empty };
return PartialView("Confirm", model);
}
Model for last step:
public class WizardModel
{
public string Phone { get; set; }
public string Name { get; set; }
}
Make sure you reference jquery.unobtrusive-ajax in your page/layout page
<script src="~/Scripts/jquery.unobtrusive-ajax.js"></script>
Wizard.cshtml
#{
ViewBag.Title = "Wizard";
Layout = "~/Views/Shared/_Layout.cshtml";
}
<h2>Wizard - Overview</h2>
#using (Ajax.BeginForm("Step1", new AjaxOptions { HttpMethod="Get", UpdateTargetId = "wizardcontainer" }))
{
<input type="submit" value="Start wizard" />
}
<div id="wizardcontainer"></div>
Step1.cshtml
<div>
<h2>Wizard - Step 1</h2>
<br />
#using(Ajax.BeginForm("Step2", new AjaxOptions { UpdateTargetId = "wizardcontainer" }))
{
#Html.Label("FullName")
#Html.TextBox("FullName")
<input type="submit" value="Next >>" />
}
</div>
Step2.cshtml
<div>
<h2>Wizard - Step 2</h2>
#using(Ajax.BeginForm("Confirm", new AjaxOptions { UpdateTargetId = "wizardcontainer" }))
{
#Html.Label("Phone")
#Html.TextBox("Phone")
#Ajax.ActionLink("<< Previous", "Step1", new AjaxOptions { UpdateTargetId = "wizardcontainer" })
<input type="submit" value="Next >>" />
}
</div>
Confirm.cshtml
#model MvcApplication2.Controllers.WizardModel
<div>
<h2>Wizard - Final Stage</h2>
Name: #Model.Name
<br />
Phone: #Model.Phone
#Ajax.ActionLink("<< Previous", "Step2", new AjaxOptions { UpdateTargetId = "wizardcontainer" })
</div>
Look here for the AjaxOnly attribute:
http://helios.ca/2009/05/27/aspnet-mvc-action-filter-ajax-only-attribute/
I must have something incorrectly setup as I can't get the UpdateModel function to correctly update my model based on information passed in via a FormCollection.
My View looks like:
#model NSLM.Models.Person
#{
ViewBag.Title = "MVC Example";
}
<h2>My MVC Model</h2>
<fieldset>
<legend>Person</legend>
#using(#Html.BeginForm())
{
<p>ID: #Html.TextBox("ID", Model.ID)</p>
<p>Forename: #Html.TextBox("Forename", Model.Forename)</p>
<p>Surname: #Html.TextBox("Surname", Model.Surname)</p>
<input type="submit" value="Submit" />
}
</fieldset>
My model is:
namespace NSLM.Models
{
public class Person
{
public int ID;
public string Forename;
public string Surname;
}
}
and my controller is:
[HttpPost]
public ActionResult Details(FormCollection collection)
{
try
{
// TODO: Add update logic here
Models.Person m = new Models.Person();
// This doesn't work i.e. the model is not updated with the form values
TryUpdateModel(m);
// This does work
int.TryParse(Request.Form["ID"], out m.ID);
m.Forename = Request.Form["Forename"];
m.Surname = Request.Form["Surname"];
return View(m);
}
catch
{
return View();
}
}
as you can see if I manually assign each property it works fine, so what have I not set that would get the model to be updated with the form values?
Thanks,
Mark
Replace fields with properties in your model, i.e.:
namespace NSLM.Models
{
public class Person
{
public int ID {get; set;}
public string Forename {get; set;}
public string Surname {get; set;}
}
}
By the time the call gets to the action method any automatic model binding has already been performed. Try changing the input parameter of your action method to accept a Person instance. In that case the model binder will try to create the instance and populate it from the values passed by your form.
try this :
view :
#model NSLM.Models.Person
#{
ViewBag.Title = "MVC Example";
}
<h2>My MVC Model</h2>
<fieldset>
<legend>Person</legend>
#using(#Html.BeginForm())
{
#Html.HiddenFor(model => model.ID)
<p>Forename: #Html.EditorFor(model => model.Name)
#Html.ValidationMessageFor(model => model.Name)
</p>
<p>Surname: #Html.EditorFor(model => model.Surname)
#Html.ValidationMessageFor(model => model.Surname)
</p>
<input type="submit" value="Submit" />
}
</fieldset>
Controller :
[HttpPost]
public ActionResult Details(Person p)
{
if (ModelState.IsValid)
{
db.Entry(p).State = EntityState.Modified;
db.SaveChanges();
return RedirectToAction("Index");
}
return View(p);
}