调用来自控制器使用ASP.NET MVC 4另一个不同的看法 [英] Calling another different view from the controller using ASP.NET MVC 4

问题描述

我有一个提交按钮的视图（Index.cshtml）。当点击提交按钮，它会调用控制器（TestController.cs）内的动作（Action01），所以在我要返回给调用者（Index.cshtml）视图中，带有自定义视图模型参数的动作结束。我该怎么做呢？

I have a view (Index.cshtml) with a submit button. When the submit button is clicked, it calls an action (Action01) within the controller (TestController.cs) so at the end of the action I want to return to the caller (Index.cshtml) view with a custom view model as a parameter. How do I do this?

在结果第一次尝试使用视图（视图名，模型）：

Results after first attempt using View("ViewName",model):

这是引发错误，因为动作控制器的测试中，所以返回时，它搜索\\查看\\测试\\指数，和我的索引页是\\查看\\首页\\目录。

An error is raised, as the action is within the controller Test, so returning, it is searching for \Views\Tests\Index, and my Index page is in \Views\Home\Index.

视图索引或它的主人没有被发现或没有视图引擎支持搜索位置。在以下地点进行了全面搜查：

The view 'Index' or its master was not found or no view engine supports the searched locations. The following locations were searched:

~/Views/Test/Index.aspx
~/Views/Test/Index.ascx
~/Views/Shared/Index.aspx
~/Views/Shared/Index.ascx
~/Views/Test/Index.cshtml
~/Views/Test/Index.vbhtml
~/Views/Shared/Index.cshtml
~/Views/Shared/Index.vbhtml

最终的解决方案：

我用返回视图（视图名，模型），我已经改变了我的目录结构，因为它是这个问题。

Final solution:

I have used return View("ViewName", model), and I have changed my directories structure as it was the problem.

推荐答案

您可以直接返回像一个不同的看法：

You can directly return a different view like:

return View("NameOfView", Model);

或者你可以做一个局部视图，并且可以返回，如：

Or you can make a partial view and can return like:

return PartialView("PartialViewName", Model);

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