如何将参数传递给ASP.NET MVC中的局部视图？

Question

假设我有这个局部视图:

Your name is <strong>@firstName @lastName</strong>

这可以通过仅限儿童的动作来访问:

[ChildActionOnly]
public ActionResult FullName(string firstName, string lastName)
{

}

我想在另一个视图中使用这个局部视图:

@Html.RenderPartial("FullName")

换句话说,我希望能够将firstName和lastName从视图传递到局部视图.我该怎么办？

Answer 1

使用此重载(RenderPartialExtensions.RenderPartial在MSDN上):

public static void RenderPartial(
    this HtmlHelper htmlHelper,
    string partialViewName,
    Object model
)

所以:

@{Html.RenderPartial(
    "FullName",
    new { firstName = model.FirstName, lastName = model.LastName});
}

Answer 2

如果您想使用ViewData,这是另一种方法:

@Html.Partial("~/PathToYourView.cshtml", null, new ViewDataDictionary { { "VariableName", "some value" } })

并检索传入的值:

@{
    string valuePassedIn = this.ViewData.ContainsKey("VariableName") ? this.ViewData["VariableName"].ToString() : string.Empty;
}

Answer 3

您需要创建一个视图模型.这样的事应该做​​......

public class FullNameViewModel
{
     public string FirstName { get; set; }
     public string LastName { get; set; }

     public FullNameViewModel() { } 

     public FullNameViewModel(string firstName, string lastName)
     {
          this.FirstName = firstName;
          this.LastName = lastName;
     }

}

然后从你的动作结果传递模型

return View("FullName", new FullNameViewModel("John", "Doe"));

并且您将能够访问@Model.FirstName并@Model.LastName相应地访问.

Answer 4

确保在Html.RenderPartial周围添加{},如下所示:

@{Html.RenderPartial("FullName", new { firstName = model.FirstName, lastName = model.LastName});}

不

@Html.RenderPartial("FullName", new { firstName = model.FirstName, lastName = model.LastName});

Answer 5

以下在dotnet 1.0.1上为我工作：

./ourView.cshtml

@Html.Partial(
  "_ourPartial.cshtml",
  new ViewDataDictionary(this.Vi??ewData) {
    {
      "hi", "hello" 
    } 
  }
);

./_ourPartial.cshtml

<h1>@this.ViewData["hi"]</h1>