控制器和视图之间传递数据 - ASP.NET MVC 4 [英] Passing data between controller and view - ASP.NET MVC 4

问题描述

我想提出一个简单的调查，为ASP.NET MVC的学习基础4.这里是我的code

I am making a simple survey, for learning basics of ASP.NET MVC 4. Here's my code

        [HttpGet]
        public ActionResult Index()
        {
            UserAndTableViewmodel Viewmodel = new UserAndTableViewmodel();
            Viewmodel.T = Deserialize();
            Viewmodel.U = new User();

            for (int i = 0; i < Viewmodel.T.Questions.Count(); i++)
            {
                Viewmodel.U.UserChoices.Add(new Choice(Viewmodel.T.Questions[i].Choices[0].Value));
            }

            return View(Viewmodel);
        }

        [HttpPost]
        public ActionResult Index(UserAndTableViewmodel Viewmodel)
        {
            // Viewmodel.T = Deserialize();

            if (ModelState.IsValid)
            {
                return View("Thanks", Viewmodel);
            }
            else
            {
                return View(Viewmodel);
            }
        }

---

中的XML code是如下：

---

The XML code is as followed:

<Table>
  <Question Content="Question one">
    <Choice Value="Answer 1" />
    <Choice Value="Answer 2" />
  </Question>
  <Question Content="Question two">
    (...)
  </Question>
</Table>

---

我传递deserialised数据索引的观点，在这里用户可以选择自己的答案。然后数据被发布到[HttpPost]我希望它渲染视图，其中其回答每一个问题时写的，但出现的问题 - Viewmodel.T 等于null 。什么是我应该做的，我不应该再反序列化呢？

---

I'm passing deserialised data to "Index" view, where user can choose his answers. Then data is post to [HttpPost] and i want it to render a view, where each question with its answer is written, but problem occurs - Viewmodel.T is equal to null. What am I supposed to do, that I shouldn't deserialize it again?

推荐答案

您不想做任何MVC4或连载deserialise。只是通过您的数据视图模型对象，将在您的视图可用。

You don't want to do any serialize or deserialise in MVC4. Just pass your data as a view model object, that will be available in your view.

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