传入RedirectToAction）使用T4MVC一个对象（

问题描述

可能重复：结果
  从操作

我在我提交表单，并根据我想重定向到一个动作的结果的看法。这correspondes并采取行动的看法是强类型的，它应该接受ResultsViewModel。

I have a view in which I submit a form and depending on the results I want to redirect to an action. The view that correspondes to and action is strongly-typed and it should accept a ResultsViewModel.

我试图通过使用T4MVC的ResultsViewModel。

I'm trying to pass the ResultsViewModel using T4MVC.

下面是code：

    [HttpPost]
    public virtual ActionResult AddEntity(string viewModel)
    {
        //Deserialize using Json.NET
        var entity = JsonConvert.DeserializeObject<MyEntity>(viewModel);

        var success = DoSomething(); //returns boolean
        if(success)
        {
            var result = new ResultsViewModel { MyEntity = entity, MessageId = 1};
            return RedirectToAction(MVC.MyController.ResultsPage(result));
        }

        var result = new ResultsViewModel { MyEntity = entity, MessageId = 2};
        return RedirectToAction(MVC.MyController.ResultsPage(result));
    }

    public virtual ActionResult ResultsPage(ResultsViewModel viewModel)
    {
        return View(viewModel);
    }

当code到达

    public virtual ActionResult ResultsPage(ResultsViewModel viewModel)
    {
        return View(viewModel);
    }

视图模型总是等于空。

viewModel is always equal to null.

我知道我可以做这样的事情：

I know I can do something like this:

return RedirectToAction("ResultsPage", new { viewModel = result });

编辑：我试图返回RedirectToAction（ResultsPage，新的{视图模型=结果}）;而我也在我的ViewModel得到一个空。

I tried the return RedirectToAction("ResultsPage", new { viewModel = result }); and I'm also getting a null in my viewModel.

不过我想找出为什么/如何使用T4MVC传递一个对象。

However I'm trying to figure out why/how to pass an object using T4MVC.

谢谢，

推荐答案

使用的TempData

[HttpPost]
public virtual ActionResult AddEntity(string viewModel)
{
    //Deserialize using Json.NET
    var entity = JsonConvert.DeserializeObject<MyEntity>(viewModel);

    var success = DoSomething(); //returns boolean
    if(success)
    {
        var result = new ResultsViewModel { MyEntity = entity, MessageId = 1};
        return RedirectToAction(MVC.MyController.ResultsPage(result));
    }

    var result = new ResultsViewModel { MyEntity = entity, MessageId = 2};

    TempData["Result"] = result;


    return RedirectToAction(MVC.MyController.ResultsPage(result));
}

public virtual ActionResult ResultsPage()
{
    ResultsViewModel model = (ResultsViewModel)TempData["Result"];
    return View(model);
}

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