我想在一个视图中有2个模型。页面包含LoginViewModel和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; }
}

我需要做另一个ViewModel持有这2个ViewModel吗?

public BigViewModel
{
    public LoginViewModel LoginViewModel{get; set;}
    public RegisterViewModel RegisterViewModel {get; set;}
}

我需要将验证属性带到视图中。这就是为什么我需要ViewModels。

有没有其他的方法,比如(没有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)
 }

当前回答

有很多方法……

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

其他回答

一个简单的方法

我们可以先呼叫所有型号

@using project.Models

然后发送你的模型与viewbag

// for list
ViewBag.Name = db.YourModel.ToList();

// for one
ViewBag.Name = db.YourModel.Find(id);

在视图中

// for list
List<YourModel> Name = (List<YourModel>)ViewBag.Name ;

//for one
YourModel Name = (YourModel)ViewBag.Name ;

然后很容易使用这个模型

使用包含多个视图模型的视图模型:

   namespace MyProject.Web.ViewModels
   {
      public class UserViewModel
      {
          public UserDto User { get; set; }
          public ProductDto Product { get; set; }
          public AddressDto Address { get; set; }
      }
   }

你认为:

  @model MyProject.Web.ViewModels.UserViewModel

  @Html.LabelFor(model => model.User.UserName)
  @Html.LabelFor(model => model.Product.ProductName)
  @Html.LabelFor(model => model.Address.StreetName)

另一种方法是使用:

@model Tuple<LoginViewModel,RegisterViewModel>

我已经在另一个例子中解释了如何在视图和控制器中使用这个方法:在ASP MVC 3的一个视图中有两个模型

在你的情况下,你可以使用下面的代码来实现它:

在视图中:

@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.

如果需要一个actionlink,你可以使用:

@Html.ActionLink("Edit", "Edit", new { id=Model.Item1.Id })

在控制器的视图方法中,需要创建一个类型为Tuple的变量,然后传递给视图。

例子:

public ActionResult Details()
{
    var tuple = new Tuple<LoginViewModel, RegisterViewModel>(new LoginViewModel(),new RegisterViewModel());
    return View(tuple);
}

或者您可以用值填充LoginViewModel和RegisterViewModel的两个实例,然后将其传递给视图。

这是一个使用IEnumerable的简化示例。

我在视图上使用了两个模型:一个是带有搜索条件的表单(SearchParams模型),另一个是用于显示结果的网格,我还纠结于如何在同一个视图上添加IEnumerable模型和另一个模型。这是我想到的,希望能帮助到一些人:

@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; }
    }
}

你总是可以在ViewBag或View Data中传递第二个对象。