504

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

这在 ASP.NET MVC beta 中可行吗？

我尝试了多个示例：

1. ASP.NET MVC Beta 中的 RenderPartial 到 String

如果我使用此示例，我会收到“发送 HTTP 标头后无法重定向。”。

2. MVC 框架：捕获视图的输出

如果我使用它，我似乎无法执行redirectToAction，因为它试图呈现一个可能不存在的视图。如果我确实返回视图，它就会完全混乱并且看起来根本不正确。

有没有人对我遇到的这些问题有任何想法/解决方案，或者对更好的问题有任何建议？

非常感谢！

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);
}

Tim Scott 接受的回答（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;
    }
}

示例用法

假设控制器调用以获取订单确认电子邮件，并传递 Site.Master 位置。

string myString = RenderViewToString(this.ControllerContext, "~/Views/Order/OrderResultEmail.aspx", "~/Views/Shared/Site.Master", this.ViewData, this.TempData);
4

16 回答 16

587

这是我想出的，它对我有用。我将以下方法添加到我的控制器基类中。（您总是可以在其他地方制作这些静态方法，我想这些方法接受控制器作为参数）

MVC2 .ascx 风格

protected string RenderViewToString<T>(string viewPath, T model) {
  ViewData.Model = model;
  using (var writer = new StringWriter()) {
    var view = new WebFormView(ControllerContext, viewPath);
    var vdd = new ViewDataDictionary<T>(model);
    var viewCxt = new ViewContext(ControllerContext, view, vdd,
                                new TempDataDictionary(), writer);
    viewCxt.View.Render(viewCxt, writer);
    return writer.ToString();
  }
}

剃刀 .cshtml 样式

public string RenderRazorViewToString(string viewName, object model)
{
  ViewData.Model = model;
  using (var sw = new StringWriter())
  {
    var viewResult = ViewEngines.Engines.FindPartialView(ControllerContext,
                                                             viewName);
    var viewContext = new ViewContext(ControllerContext, viewResult.View,
                                 ViewData, TempData, sw);
    viewResult.View.Render(viewContext, sw);
    viewResult.ViewEngine.ReleaseView(ControllerContext, viewResult.View);
    return sw.GetStringBuilder().ToString();
  }
}

编辑：添加 Razor 代码。

于 2010-05-03T17:26:35.870 回答
71

这个答案不在我的路上。这最初来自https://stackoverflow.com/a/2759898/2318354 但在这里我展示了将它与“静态”关键字一起使用以使其对所有控制器通用的方法。

为此，您必须static在类文件中创建类。（假设您的类文件名是 Utils.cs ）

此示例适用于 Razor。

实用程序.cs

public static class RazorViewToString
{
    public static string RenderRazorViewToString(this Controller controller, string viewName, object model)
    {
        controller.ViewData.Model = model;
        using (var sw = new StringWriter())
        {
            var viewResult = ViewEngines.Engines.FindPartialView(controller.ControllerContext, viewName);
            var viewContext = new ViewContext(controller.ControllerContext, viewResult.View, controller.ViewData, controller.TempData, sw);
            viewResult.View.Render(viewContext, sw);
            viewResult.ViewEngine.ReleaseView(controller.ControllerContext, viewResult.View);
            return sw.GetStringBuilder().ToString();
        }
    }
}

现在，您可以通过将“this”作为参数传递给控制器​​，在控制器文件中添加名称空间，从而从控制器调用此类。

string result = RazorViewToString.RenderRazorViewToString(this ,"ViewName", model);

正如@Sergey 给出的建议，这个扩展方法也可以从控制器调用，如下所示

string result = this.RenderRazorViewToString("ViewName", model);

我希望这对您使代码干净整洁有用。

于 2013-09-24T09:42:01.873 回答
33

这对我有用：

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;
    }
}
于 2009-01-27T19:48:12.297 回答
32

我找到了一个新的解决方案，它可以将视图呈现为字符串，而不必弄乱当前 HttpContext 的响应流（它不允许您更改响应的 ContentType 或其他标头）。

基本上，你所做的就是创建一个假的 HttpContext 来让视图自己呈现：

/// <summary>Renders a view to string.</summary>
public static string RenderViewToString(this Controller controller,
                                        string viewName, object viewData) {
    //Create memory writer
    var sb = new StringBuilder();
    var memWriter = new StringWriter(sb);

    //Create fake http context to render the view
    var fakeResponse = new HttpResponse(memWriter);
    var fakeContext = new HttpContext(HttpContext.Current.Request, fakeResponse);
    var fakeControllerContext = new ControllerContext(
        new HttpContextWrapper(fakeContext),
        controller.ControllerContext.RouteData,
        controller.ControllerContext.Controller);

    var oldContext = HttpContext.Current;
    HttpContext.Current = fakeContext;

    //Use HtmlHelper to render partial view to fake context
    var html = new HtmlHelper(new ViewContext(fakeControllerContext,
        new FakeView(), new ViewDataDictionary(), new TempDataDictionary()),
        new ViewPage());
    html.RenderPartial(viewName, viewData);

    //Restore context
    HttpContext.Current = oldContext;    

    //Flush memory and return output
    memWriter.Flush();
    return sb.ToString();
}

/// <summary>Fake IView implementation used to instantiate an HtmlHelper.</summary>
public class FakeView : IView {
    #region IView Members

    public void Render(ViewContext viewContext, System.IO.TextWriter writer) {
        throw new NotImplementedException();
    }

    #endregion
}

这适用于 ASP.NET MVC 1.0，以及 ContentResult、JsonResult 等（更改原始 HttpResponse 上的标头不会引发“服务器无法在发送 HTTP 标头后设置内容类型”异常）。

更新：在 ASP.NET MVC 2.0 RC 中，代码发生了一些变化，因为我们必须传入StringWriter用于将视图写入的ViewContext:

//...

//Use HtmlHelper to render partial view to fake context
var html = new HtmlHelper(
    new ViewContext(fakeControllerContext, new FakeView(),
        new ViewDataDictionary(), new TempDataDictionary(), memWriter),
    new ViewPage());
html.RenderPartial(viewName, viewData);

//...
于 2009-08-06T20:32:57.443 回答
13

本文介绍了不同场景下如何将 View 渲染为字符串：

  1. MVC 控制器调用另一个自己的 ActionMethods
  2. MVC 控制器调用另一个 MVC 控制器的 ActionMethod
  3. 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GitHub 上 Rick Stahl 的 WestwindToolkit 的一部分。

用法（3. - WebAPI 示例）：

string html = ViewRenderer.RenderView("~/Areas/ReportDetail/Views/ReportDetail/Index.cshtml", ReportVM.Create(id));
于 2014-10-02T07:13:55.033 回答
10

如果你想完全放弃 MVC，从而避免所有的 HttpContext 混乱......

using RazorEngine;
using RazorEngine.Templating; // For extension methods.

string razorText = System.IO.File.ReadAllText(razorTemplateFileLocation);
string emailBody = Engine.Razor.RunCompile(razorText, "templateKey", typeof(Model), model);

这里使用了很棒的开源 Razor 引擎： https ://github.com/Antaris/RazorEngine

于 2013-09-18T19:49:01.640 回答
7

ASP NET CORE 的附加提示：

界面：

public interface IViewRenderer
{
  Task<string> RenderAsync<TModel>(Controller controller, string name, TModel model);
}

执行：

public class ViewRenderer : IViewRenderer
{
  private readonly IRazorViewEngine viewEngine;

  public ViewRenderer(IRazorViewEngine viewEngine) => this.viewEngine = viewEngine;

  public async Task<string> RenderAsync<TModel>(Controller controller, string name, TModel model)
  {
    ViewEngineResult viewEngineResult = this.viewEngine.FindView(controller.ControllerContext, name, false);

    if (!viewEngineResult.Success)
    {
      throw new InvalidOperationException(string.Format("Could not find view: {0}", name));
    }

    IView view = viewEngineResult.View;
    controller.ViewData.Model = model;

    await using var writer = new StringWriter();
    var viewContext = new ViewContext(
       controller.ControllerContext,
       view,
       controller.ViewData,
       controller.TempData,
       writer,
       new HtmlHelperOptions());

       await view.RenderAsync(viewContext);

       return writer.ToString();
  }
}

登记在Startup.cs

...
 services.AddSingleton<IViewRenderer, ViewRenderer>();
...

以及在控制器中的用法：

public MyController: Controller
{
  private readonly IViewRenderer renderer;
  public MyController(IViewRendere renderer) => this.renderer = renderer;
  public async Task<IActionResult> MyViewTest
  {
    var view = await this.renderer.RenderAsync(this, "MyView", model);
    return new OkObjectResult(view);
  }
}
于 2019-09-28T08:19:18.980 回答
5

您可以使用这种方式获取字符串中的视图

protected string RenderPartialViewToString(string viewName, object model)
{
    if (string.IsNullOrEmpty(viewName))
        viewName = ControllerContext.RouteData.GetRequiredString("action");

    if (model != null)
        ViewData.Model = model;

    using (StringWriter sw = new StringWriter())
    {
        ViewEngineResult viewResult = ViewEngines.Engines.FindPartialView(ControllerContext, viewName);
        ViewContext viewContext = new ViewContext(ControllerContext, viewResult.View, ViewData, TempData, sw);
        viewResult.View.Render(viewContext, sw);

        return sw.GetStringBuilder().ToString();
    }
}

我们以两种方式调用此方法

string strView = RenderPartialViewToString("~/Views/Shared/_Header.cshtml", null)

或者

var model = new Person()
string strView = RenderPartialViewToString("~/Views/Shared/_Header.cshtml", model)
于 2015-03-23T12:47:24.880 回答
5

要在服务层中将视图呈现给字符串而不必传递 ControllerContext，这里有一篇很好的 Rick Strahl 文章http://www.codemag.com/Article/1312081，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;
}

然后在服务类中渲染视图：

var stringView = RenderViewToString(CreateController<GenericController>().ControllerContext, "~/Path/To/View/Location/_viewName.cshtml", theViewModel, true);
于 2017-03-02T15:32:05.603 回答
3

我正在使用 MVC 1.0 RTM，但上述解决方案都不适合我。但是这个做到了：

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
于 2009-07-13T20:44:07.033 回答
2

我从另一个网站看到了 MVC 3 和 Razor 的实现，它对我有用：

    public static string RazorRender(Controller context, string DefaultAction)
    {
        string Cache = string.Empty;
        System.Text.StringBuilder sb = new System.Text.StringBuilder();
        System.IO.TextWriter tw = new System.IO.StringWriter(sb); 

        RazorView view_ = new RazorView(context.ControllerContext, DefaultAction, null, false, null);
        view_.Render(new ViewContext(context.ControllerContext, view_, new ViewDataDictionary(), new TempDataDictionary(), tw), tw);

        Cache = sb.ToString(); 

        return Cache;

    } 

    public static string RenderRazorViewToString(string viewName, object model)
    {

        ViewData.Model = model;
        using (var sw = new StringWriter())
        {
            var viewResult = ViewEngines.Engines.FindPartialView(ControllerContext, viewName);
            var viewContext = new ViewContext(ControllerContext, viewResult.View, ViewData, TempData, sw);
            viewResult.View.Render(viewContext, sw);
            return sw.GetStringBuilder().ToString();
        }
    } 

    public static class HtmlHelperExtensions
    {
        public static string RenderPartialToString(ControllerContext context, string partialViewName, ViewDataDictionary viewData, TempDataDictionary tempData)
        {
            ViewEngineResult result = ViewEngines.Engines.FindPartialView(context, partialViewName);

            if (result.View != null)
            {
                StringBuilder sb = new StringBuilder();
                using (StringWriter sw = new StringWriter(sb))
                {
                    using (HtmlTextWriter output = new HtmlTextWriter(sw))
                    {
                        ViewContext viewContext = new ViewContext(context, result.View, viewData, tempData, output);
                        result.View.Render(viewContext, output);
                    }
                }
                return sb.ToString();
            } 

            return String.Empty;

        }

    }

更多关于Razor 渲染 - MVC3 视图渲染到字符串

于 2012-02-20T21:34:56.970 回答
1

小建议

对于强类型模型，只需将其添加到 ViewData.Model 属性，然后再传递给 RenderViewToString。例如

this.ViewData.Model = new OrderResultEmailViewModel(order);
string myString = RenderViewToString(this.ControllerContext, "~/Views/Order/OrderResultEmail.aspx", "~/Views/Shared/Site.Master", this.ViewData, this.TempData);
于 2009-06-15T16:42:27.087 回答
0

要重复一个更未知的问题，请查看MvcIntegrationTestFramework。

它可以节省您编写自己的助手来流式传输结果，并且被证明可以很好地工作。我假设这将在一个测试项目中，作为奖励，一旦你有了这个设置，你就会拥有其他测试功能。主要麻烦可能是整理依赖链。

 private static readonly string mvcAppPath = 
     Path.GetFullPath(AppDomain.CurrentDomain.BaseDirectory 
     + "\\..\\..\\..\\MyMvcApplication");
 private readonly AppHost appHost = new AppHost(mvcAppPath);

    [Test]
    public void Root_Url_Renders_Index_View()
    {
        appHost.SimulateBrowsingSession(browsingSession => {
            RequestResult result = browsingSession.ProcessRequest("");
            Assert.IsTrue(result.ResponseText.Contains("<!DOCTYPE html"));
        });
}
于 2010-03-31T15:58:05.017 回答
0

这是我为 ASP.NETCore RC2 编写的一个类。我使用它，所以我可以使用 Razor 生成 html 电子邮件。

using Microsoft.AspNetCore.Http;
using Microsoft.AspNetCore.Mvc;
using Microsoft.AspNetCore.Mvc.Abstractions;
using Microsoft.AspNetCore.Mvc.ModelBinding;
using Microsoft.AspNetCore.Mvc.Rendering;
using Microsoft.AspNetCore.Mvc.ViewEngines;
using Microsoft.AspNetCore.Mvc.ViewFeatures;
using Microsoft.AspNetCore.Routing;
using System.IO;
using System.Threading.Tasks;

namespace cloudscribe.Web.Common.Razor
{
    /// <summary>
    /// the goal of this class is to provide an easy way to produce an html string using 
    /// Razor templates and models, for use in generating html email.
    /// </summary>
    public class ViewRenderer
    {
        public ViewRenderer(
            ICompositeViewEngine viewEngine,
            ITempDataProvider tempDataProvider,
            IHttpContextAccessor contextAccesor)
        {
            this.viewEngine = viewEngine;
            this.tempDataProvider = tempDataProvider;
            this.contextAccesor = contextAccesor;
        }

        private ICompositeViewEngine viewEngine;
        private ITempDataProvider tempDataProvider;
        private IHttpContextAccessor contextAccesor;

        public async Task<string> RenderViewAsString<TModel>(string viewName, TModel model)
        {

            var viewData = new ViewDataDictionary<TModel>(
                        metadataProvider: new EmptyModelMetadataProvider(),
                        modelState: new ModelStateDictionary())
            {
                Model = model
            };

            var actionContext = new ActionContext(contextAccesor.HttpContext, new RouteData(), new ActionDescriptor());
            var tempData = new TempDataDictionary(contextAccesor.HttpContext, tempDataProvider);

            using (StringWriter output = new StringWriter())
            {

                ViewEngineResult viewResult = viewEngine.FindView(actionContext, viewName, true);

                ViewContext viewContext = new ViewContext(
                    actionContext,
                    viewResult.View,
                    viewData,
                    tempData,
                    output,
                    new HtmlHelperOptions()
                );

                await viewResult.View.RenderAsync(viewContext);

                return output.GetStringBuilder().ToString();
            }
        }
    }
}
于 2016-06-09T11:28:41.753 回答
0

当我使用上述方法出错时，我找到了一种更好的方法来呈现剃刀视图页面，这个解决方案适用于 web 表单环境和 mvc 环境。不需要控制器。

这是代码示例，在此示例中，我使用异步 http 处理程序模拟了 mvc 操作：

    /// <summary>
    /// Enables processing of HTTP Web requests asynchronously by a custom HttpHandler that implements the IHttpHandler interface.
    /// </summary>
    /// <param name="context">An HttpContext object that provides references to the intrinsic server objects.</param>
    /// <returns>The task to complete the http request.</returns>
    protected override async Task ProcessRequestAsync(HttpContext context)
    {
        if (this._view == null)
        {
            this.OnError(context, new FileNotFoundException("Can not find the mvc view file.".Localize()));
            return;
        }
        object model = await this.LoadModelAsync(context);
        WebPageBase page = WebPageBase.CreateInstanceFromVirtualPath(this._view.VirtualPath);
        using (StringWriter sw = new StringWriter())
        {
            page.ExecutePageHierarchy(new WebPageContext(new HttpContextWrapper(context), page, model), sw);
            await context.Response.Output.WriteAsync(sw.GetStringBuilder().ToString());
        }
    }
于 2016-08-12T16:33:38.907 回答
0

对我来说最简单的方法是：

  public string GetFileAsString(string path)
  {
        var html = "";                        

        FileStream fileStream = new FileStream(path, FileMode.Open);

        using (StreamReader reader = new StreamReader(fileStream))
        {
            html += reader.ReadLine();
        }

        return html;
   }

我将它用于电子邮件并确保文件仅包含 CSS 和 HTML

于 2021-11-19T17:59:21.110 回答