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)
 }

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

我建议使用Html。RenderAction和PartialViewResults来完成这个;它将允许您显示相同的数据,但每个部分视图仍然有一个单一的视图模型,并消除了对BigViewModel的需要

你的视图包含如下内容:

@Html.RenderAction("Login")
@Html.RenderAction("Register")

其中Login和Register都是你控制器中的动作,定义如下:

public PartialViewResult Login( )
{
    return PartialView( "Login", new LoginViewModel() );
}

public PartialViewResult Register( )
{
    return PartialView( "Register", new RegisterViewModel() );
}

Login & Register将是位于当前View文件夹或共享文件夹中的用户控件,并像这样:

/views/Shared/Login.cshtml: (或 /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: (或 /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)
}

你有一个单独的控制器动作,每个动作都有视图和视图文件每个动作都完全不同,不依赖于其他任何东西。

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。

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

我想说的是,我的解决方案就像这个stackoverflow页面上提供的答案:ASP。NET MVC 4,多个模型在一个视图?

然而,在我的情况下,他们在他们的控制器中使用的linq查询不适合我。

这就是所谓的查询:

var viewModels = 
        (from e in db.Engineers
         select new MyViewModel
         {
             Engineer = e,
             Elements = e.Elements,
         })
        .ToList();

因此,“在您的视图中,只需指定您正在使用视图模型的集合”对我来说也不适用。

然而,稍微改变一下这个解决方案确实对我有用。这是我的解决方案,希望对大家有所帮助。

这是我的视图模型,我知道我将只有一个团队,但该团队可能有多个板(顺便说一下,我在我的Models文件夹中有一个ViewModels文件夹,因此有命名空间):

namespace TaskBoard.Models.ViewModels
{
    public class TeamBoards
    {
        public Team Team { get; set; }
        public List<Board> Boards { get; set; }
    }
}

这是我的控制器。这是与上面引用的链接中的解决方案最显著的区别。我构建了ViewModel以不同的方式发送到视图。

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);
        }

在我看来,我没有将它指定为列表。我只做@model taskboard。models。viewmodels。“然后,当我迭代团队的板时,我只需要每个。以下是我的观点:

@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>

我对ASP还是个新手。NET MVC,所以我花了一点时间来弄清楚。所以,我希望这篇文章能帮助人们在更短的时间内为他们的项目找到答案。: -)

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)

在你的模型和LoginViewModel和RegisterViewModel属性中创建一个新类: 公共类UserDefinedModel() { 属性a1作为LoginViewModel property a2 as RegisterViewModel } 然后在视图中使用UserDefinedModel。

将这个ModelCollection.cs添加到您的模型中

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)];
   }
 }
}

控制器:

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);
  }
}

视图:

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

我是否需要创建另一个包含这两个视图的视图?

回答:不

有没有其他的方法,比如(没有BigViewModel):

是的,你可以使用Tuple(在有多个模型的视图中带来魔力)。

代码:

 @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)
     }

你总是可以在ViewBag或View Data中传递第二个对象。

我的建议是创建一个大的视图模型:

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

在你的索引。Cshtml,如果你有两个部分:

@addTagHelper *,Microsoft.AspNetCore.Mvc.TagHelpers
@model .BigViewModel

@await Html.PartialAsync("_LoginViewPartial", Model.LoginViewModel)

@await Html.PartialAsync("_RegisterViewPartial ", Model.RegisterViewModel )

在控制器中:

model=new BigViewModel();
model.LoginViewModel=new LoginViewModel();
model.RegisterViewModel=new RegisterViewModel(); 

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; }
    }
}