如何渲染一个ASP。NET MVC视图作为字符串?

我想输出两个不同的视图(一个作为字符串，将作为电子邮件发送)，另一个是显示给用户的页面。

这在ASP中可能吗?NET MVC beta版?

我试过很多例子:

1. 在ASP。NET MVC Beta版

如果我使用这个例子，我会收到“HTTP后不能重定向” 报头已发送。”

2. MVC框架:捕获视图的输出

如果我使用这个，我似乎无法做一个重定向toaction，因为它试图呈现一个可能不存在的视图。如果我返回视图完全是一团糟，看起来一点都不对。

有人对我提出的这些问题有什么想法/解决方案吗?或者有什么更好的建议吗?

很多谢谢!

下面是一个例子。我要做的是创建GetViewForEmail方法:

public ActionResult OrderResult(string ref)
{
    //Get the order
    Order order = OrderService.GetOrder(ref);

    //The email helper would do the meat and veg by getting the view as a string
    //Pass the control name (OrderResultEmail) and the model (order)
    string emailView = GetViewForEmail("OrderResultEmail", order);

    //Email the order out
    EmailHelper(order, emailView);
    return View("OrderResult", order);
}

接受蒂姆·斯科特的回答(我做了一些修改和格式化):

public virtual string RenderViewToString(
    ControllerContext controllerContext,
    string viewPath,
    string masterPath,
    ViewDataDictionary viewData,
    TempDataDictionary tempData)
{
    Stream filter = null;
    ViewPage viewPage = new ViewPage();

    //Right, create our view
    viewPage.ViewContext = new ViewContext(controllerContext, new WebFormView(viewPath, masterPath), viewData, tempData);

    //Get the response context, flush it and get the response filter.
    var response = viewPage.ViewContext.HttpContext.Response;
    response.Flush();
    var oldFilter = response.Filter;

    try
    {
        //Put a new filter into the response
        filter = new MemoryStream();
        response.Filter = filter;

        //Now render the view into the memorystream and flush the response
        viewPage.ViewContext.View.Render(viewPage.ViewContext, viewPage.ViewContext.HttpContext.Response.Output);
        response.Flush();

        //Now read the rendered view.
        filter.Position = 0;
        var reader = new StreamReader(filter, response.ContentEncoding);
        return reader.ReadToEnd();
    }
    finally
    {
        //Clean up.
        if (filter != null)
        {
            filter.Dispose();
        }

        //Now replace the response filter
        response.Filter = oldFilter;
    }
}

示例使用

假设控制器通过Site调用订单确认电子邮件。主人的位置。

string myString = RenderViewToString(this.ControllerContext, "~/Views/Order/OrderResultEmail.aspx", "~/Views/Shared/Site.Master", this.ViewData, this.TempData);

当前回答

这对我来说很管用:

public virtual string RenderView(ViewContext viewContext)
{
    var response = viewContext.HttpContext.Response;
    response.Flush();
    var oldFilter = response.Filter;
    Stream filter = null;
    try
    {
        filter = new MemoryStream();
        response.Filter = filter;
        viewContext.View.Render(viewContext, viewContext.HttpContext.Response.Output);
        response.Flush();
        filter.Position = 0;
        var reader = new StreamReader(filter, response.ContentEncoding);
        return reader.ReadToEnd();
    }
    finally
    {
        if (filter != null)
        {
            filter.Dispose();
        }
        response.Filter = oldFilter;
    }
}

2009-01-27 19:48:12

其他回答

这对我来说很管用:

public virtual string RenderView(ViewContext viewContext)
{
    var response = viewContext.HttpContext.Response;
    response.Flush();
    var oldFilter = response.Filter;
    Stream filter = null;
    try
    {
        filter = new MemoryStream();
        response.Filter = filter;
        viewContext.View.Render(viewContext, viewContext.HttpContext.Response.Output);
        response.Flush();
        filter.Position = 0;
        var reader = new StreamReader(filter, response.ContentEncoding);
        return reader.ReadToEnd();
    }
    finally
    {
        if (filter != null)
        {
            filter.Dispose();
        }
        response.Filter = oldFilter;
    }
}

2009-01-27 19:48:12

为了在服务层中呈现一个字符串的视图，而不需要传递ControllerContext, Rick Strahl在http://www.codemag.com/Article/1312081有一篇很好的文章，它创建了一个通用控制器。代码摘要如下:

// Some Static Class
public static string RenderViewToString(ControllerContext context, string viewPath, object model = null, bool partial = false)
{
    // first find the ViewEngine for this view
    ViewEngineResult viewEngineResult = null;
    if (partial)
        viewEngineResult = ViewEngines.Engines.FindPartialView(context, viewPath);
    else
        viewEngineResult = ViewEngines.Engines.FindView(context, viewPath, null);

    if (viewEngineResult == null)
        throw new FileNotFoundException("View cannot be found.");

    // get the view and attach the model to view data
    var view = viewEngineResult.View;
    context.Controller.ViewData.Model = model;

    string result = null;

    using (var sw = new StringWriter())
    {
        var ctx = new ViewContext(context, view, context.Controller.ViewData, context.Controller.TempData, sw);
        view.Render(ctx, sw);
        result = sw.ToString();
    }

    return result;
}

// In the Service Class
public class GenericController : Controller
{ }

public static T CreateController<T>(RouteData routeData = null) where T : Controller, new()
{
    // create a disconnected controller instance
    T controller = new T();

    // get context wrapper from HttpContext if available
    HttpContextBase wrapper;
    if (System.Web.HttpContext.Current != null)
        wrapper = new HttpContextWrapper(System.Web.HttpContext.Current);
    else
        throw new InvalidOperationException("Cannot create Controller Context if no active HttpContext instance is available.");

    if (routeData == null)
        routeData = new RouteData();

    // add the controller routing if not existing
    if (!routeData.Values.ContainsKey("controller") &&
        !routeData.Values.ContainsKey("Controller"))
        routeData.Values.Add("controller", controller.GetType().Name.ToLower().Replace("controller", ""));

    controller.ControllerContext = new ControllerContext(wrapper, routeData, controller);
    return controller;
}

然后在Service类中呈现View:

var stringView = RenderViewToString(CreateController<GenericController>().ControllerContext, "~/Path/To/View/Location/_viewName.cshtml", theViewModel, true);

2017-03-02 15:32:05

快速提示

对于强类型模型，只需将其添加到ViewData。Model属性，然后传递给RenderViewToString。如

this.ViewData.Model = new OrderResultEmailViewModel(order);
string myString = RenderViewToString(this.ControllerContext, "~/Views/Order/OrderResultEmail.aspx", "~/Views/Shared/Site.Master", this.ViewData, this.TempData);

2009-06-15 16:42:27

我使用的是MVC 1.0 RTM，上面的解决方案都不适合我。但这一个做到了:

Public Function RenderView(ByVal viewContext As ViewContext) As String

    Dim html As String = ""

    Dim response As HttpResponse = HttpContext.Current.Response

    Using tempWriter As New System.IO.StringWriter()

        Dim privateMethod As MethodInfo = response.GetType().GetMethod("SwitchWriter", BindingFlags.NonPublic Or BindingFlags.Instance)

        Dim currentWriter As Object = privateMethod.Invoke(response, BindingFlags.NonPublic Or BindingFlags.Instance Or BindingFlags.InvokeMethod, Nothing, New Object() {tempWriter}, Nothing)

        Try
            viewContext.View.Render(viewContext, Nothing)
            html = tempWriter.ToString()
        Finally
            privateMethod.Invoke(response, BindingFlags.NonPublic Or BindingFlags.Instance Or BindingFlags.InvokeMethod, Nothing, New Object() {currentWriter}, Nothing)
        End Try

    End Using

    Return html

End Function

2009-07-13 20:44:07

这篇文章描述了如何在不同的场景中呈现一个视图给字符串:

MVC控制器调用它自己的另一个actionmethod MVC控制器调用另一个MVC控制器的ActionMethod WebAPI控制器调用MVC控制器的ActionMethod

解决方案/代码作为一个名为ViewRenderer的类提供。它是Rick Stahl在GitHub上的WestwindToolkit的一部分。

使用(3。- WebAPI示例):

string html = ViewRenderer.RenderView("~/Areas/ReportDetail/Views/ReportDetail/Index.cshtml", ReportVM.Create(id));

2014-10-02 07:13:55

如何渲染一个ASP。NET MVC视图作为字符串?

推荐文章

最新文章

标签