MVC 3不能传递string作为视图的模型？

我有一个奇怪的问题，我的模型传递给视图

调节器

[Authorize] public ActionResult Sth() { return View("~/Views/Sth/Sth.cshtml", "abc"); } 

视图

 @model string @{ ViewBag.Title = "lorem"; Layout = "~/Views/Shared/Default.cshtml"; } 

错误消息

 The view '~/Views/Sth/Sth.cshtml' or its master was not found or no view engine supports the searched locations. The following locations were searched: ~/Views/Sth/Sth.cshtml ~/Views/Sth/abc.master //string model is threated as a possible Layout's name ? ~/Views/Shared/abc.master ~/Views/Sth/abc.cshtml ~/Views/Sth/abc.vbhtml ~/Views/Shared/abc.cshtml ~/Views/Shared/abc.vbhtml 

为什么我不能传递一个简单的string作为模型？

Answer 1

是的，你可以如果你正在使用正确的超载 ：

 return View("~/Views/Sth/Sth.cshtml" /* view name*/, null /* master name */, "abc" /* model */); 

Answer 2

如果使用命名参数，则可以省略完全给出第一个参数

 return View(model:"abc"); 

要么

 return View(viewName:"~/Views/Sth/Sth.cshtml", model:"abc"); 

也将达到目的。

Answer 3

你的意思是这个View过载：

 protected internal ViewResult View(string viewName, Object model) 

MVC被这个过载所困惑：

 protected internal ViewResult View(string viewName, string masterName) 

使用这个过载：

 protected internal virtual ViewResult View(string viewName, string masterName, Object model) 

这条路：

 return View("~/Views/Sth/Sth.cshtml", null , "abc"); 

顺便说一下，你可以使用这个：

 return View("Sth", null, "abc"); 

在MSDN上重载parsing

Answer 4

如果将该string声明为对象，它也可以工作：

 object str = "abc"; return View(str); 

要么：

 return View("abc" as object); 

Answer 5

如果您为前两个parameter passingnull，它也可以工作：

 return View(null, null, "abc");