c# - 从 ActionFilterAttribute 返回带有模型的 View

Question

当在强类型 View 上使用内置验证助手实现错误处理时，您通常会在 Controller 中创建一个 try/catch block ，并返回一个 View 及其对应的模型作为 的参数View() 方法:


Controller

public class MessageController : Controller
{
    [AcceptVerbs(HttpVerbs.Post)]
    public ActionResult Create(Models.Entities.Message message)
    {
        try
        {
            // Insert model into database
            var dc = new DataContext();
            dc.Messages.InsertOnSubmit(message);
            dc.SubmitChanges();

            return RedirectToAction("List");
        }
        catch
        {
            /* If insert fails, return a view with it's corresponding model to
               enable validation helpers */
            return View(message);
        }
    }
}

观点

<%@ Page
    Language="C#"
    Inherits="System.Web.Mvc.ViewPage<Models.Entities.Message>" %>

<%= Html.ValidationSummary("Fill out fields marked with *") %>

<% using (Html.BeginForm()) { %>

    <div><%= Html.TextBox("MessageText") %></div>

    <div><%= Html.ValidationMessage("MessageText", "*") %></div>

<% } %>

我已经以 ActionFilterAttribute 的形式实现了一个简单的错误处理程序，它将能够重定向到一个通用错误 View ，或者重定向到抛出异常的 View ，并让验证助手活跃起来。

这是我的 ActionFilterAttribute 的样子:

public class ErrorLoggingAttribute : ActionFilterAttribute, IExceptionFilter
{
    private Boolean _onErrorRedirectToGenericErrorView;

    /// <param name="onErrorRedirectToGenericErrorView">
    /// True: redirect to a generic error view.
    /// False: redirect back the view which threw an exception
    /// </param>
    public ErrorLoggingAttribute(Boolean onErrorRedirectToGenericErrorView)
    {
        _onErrorRedirectToGenericErrorView = onErrorRedirectToGenericErrorView;
    }

    public void OnException(ExceptionContext ec)
    {
        if (_onErrorRedirectToGenericErrorView)
        {
            /* Redirect back to the view where the exception was thrown and
               include it's model so the validation helpers will work */
        }
        else
        {
            // Redirect to a generic error view
            ec.Result = new RedirectToRouteResult(new RouteValueDictionary
            {
                {"controller", "Error"},
                {"action", "Index"}
            });

            ec.ExceptionHandled = true;
        }
    }
}

重定向到抛出异常的 View 相当简单。但问题在于:为了让验证助手工作，您需要为 View 提供它的模型。

如何返回抛出异常的 View 并为 View 提供相应的模型？ (在本例中 Models.Entities.Message)。

Answer 1

我让它工作了!

出于某些奇怪的原因，我需要做的就是将 ViewData 传递给新的 ResultView。

完整代码如下:

public class ErrorLoggingAttribute : ActionFilterAttribute, IExceptionFilter
{
    private String _controllerName, _actionName;
    private Boolean _redirectToGenericView = false;


    public ErrorLoggingAttribute()
    {
    }


    public ErrorLoggingAttribute(String actionName, String controllerName)
    {
        _controllerName = controllerName;
        _actionName = actionName;
        _redirectToGenericView = true;
    }


    void IExceptionFilter.OnException(ExceptionContext ec)
    {
        // log error

        if (_redirectToGenericView)
        {
            ec.Result = new RedirectToRouteResult(new RouteValueDictionary
            {
                {"controller", _controllerName},
                {"action", _actionName}
            });
        }
        else
        {
            ec.Result = new ViewResult
            {
                ViewName = ((RouteData) ec.RouteData).Values["action"].ToString(),
                TempData = ec.Controller.TempData,
                ViewData = ec.Controller.ViewData
            };
        }

        ec.ExceptionHandled = true;
    }
}

用法

以下是如何在 Controller 操作上使用该属性，以重定向到相同的 View (与它是关联的模型)以在发生异常时启用标准验证助手:

[ErrorLogging]
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult Create(Models.Entities.Message message)
{
    var dc = new Models.DataContext();
    dc.Messages.InsertOnSubmit(message);
    dc.SubmitChanges();

    return RedirectToAction("List", new { id = message.MessageId });
}

下面是发生异常时如何使用该属性重定向到通用 View :

[ErrorLogging("ControllerName", "ViewName")]
[AcceptVerbs(HttpVerbs.Post)]
public ActionResult Create(Models.Entities.Message message)

这是逻辑的完全分离。 Controller 中只有最基本的内容。