在 OnException(ExceptionContext filterContext) 内部时，有什么方法可以恢复传递给 POST 操作的模型?

问题描述

情况是这样的:

我找不到获取传递给 POST 操作方法的 viewModel 的方法.

I can't find a way of getting the viewModel that was passed to the POST action method.

[HttpPost]
public ActionResult Edit(SomeCoolModel viewModel)
{
    // Some Exception happens here during the action execution...
}

在控制器可覆盖的OnException中:

protected override void OnException(ExceptionContext filterContext)
{
    ...

    filterContext.Result = new ViewResult
    {
        ViewName = filterContext.RouteData.Values["action"].ToString(),
        TempData = filterContext.Controller.TempData,
        ViewData = filterContext.Controller.ViewData
    };
}

调试代码时filterContext.Controller.ViewData为null，因为在代码执行时发生异常并且没有返回视图.

When debugging the code filterContext.Controller.ViewData is null since the exception occurred while the code was executing and no view was returned.

无论如何，我看到 filterContext.Controller.ViewData.ModelState 已填充并具有我需要的所有值，但我没有完整的 ViewData =>viewModel 对象可用.:(

Anyways I see that filterContext.Controller.ViewData.ModelState is filled and has all the values that I need but I don't have the full ViewData => viewModel object available. :(

我想将与发布的 data/ViewModel 相同的 View 在一个中心点返回给用户.希望你能明白我的意思.

I want to return the same View with the posted data/ViewModel back to the user in a central point. Hope you get my drift.

是否还有其他途径可以实现目标?

Is there any other path I can follow to achieve the objective?

推荐答案

您可以创建一个从 DefaultModelBinder 并将模型分配给 TempData:

You could create a custom model binder that inherits from DefaultModelBinder and assign the model to TempData:

public class MyCustomerBinder : DefaultModelBinder
{
    protected override void OnModelUpdated(ControllerContext controllerContext, ModelBindingContext bindingContext)
    {
        base.OnModelUpdated(controllerContext, bindingContext);

        controllerContext.Controller.TempData["model"] = bindingContext.Model;
    }
}

并在Global.asax中注册:

ModelBinders.Binders.DefaultBinder = new MyCustomerBinder();

然后访问它:

protected override void OnException(ExceptionContext filterContext)
{
    var model = filterContext.Controller.TempData["model"];

    ...
}

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