从动作写入输出流

Question

由于一些奇怪的原因，我想直接从控制器操作将HTML写入Response流。 （我理解MVC分离，但这是一个特殊情况。）

我可以直接写入HttpResponse流吗？在那种情况下，控制器动作应该返回哪个IView对象？我可以返回'空'吗？

Answer 1

你可以做return Content(...);的地方，如果我没有记错，...将要直接写入到输出流，或者什么都没有的东西。

看看在Controller的Content方法：http://aspnet.codeplex.com/SourceControl/changeset/view/22907#266451

而且ContentResult：http://aspnet.codeplex.com/SourceControl/changeset/view/22907#266450

Answer 2

是的，你可以直接写入响应。完成后，您可以调用CompleteRequest（），并且不需要返回任何内容。

例如：

// GET: /Test/Edit/5 
public ActionResult Edit(int id) 
{ 

    Response.Write("hi"); 
    HttpContext.ApplicationInstance.CompleteRequest(); 

    return View();  // does not execute! 
} 

Answer 3

写自己的操作结果。这里的矿井中的一个的示例：

public class RssResult : ActionResult 
{ 
    public RssFeed RssFeed { get; set; } 

    public RssResult(RssFeed feed) { 
     RssFeed = feed; 
    } 

    public override void ExecuteResult(ControllerContext context) { 
     context.HttpContext.Response.ContentType = "application/rss+xml"; 
     SyndicationResourceSaveSettings settings = new SyndicationResourceSaveSettings(); 
     settings.CharacterEncoding = new UTF8Encoding(false); 
     RssFeed.Save(context.HttpContext.Response.OutputStream, settings); 
    } 
} 

Answer 4

我用从FileResult派生的类来实现这一点使用正常MVC模式：

/// <summary> 
/// MVC action result that generates the file content using a delegate that writes the content directly to the output stream. 
/// </summary> 
public class FileGeneratingResult : FileResult 
{ 
    /// <summary> 
    /// The delegate that will generate the file content. 
    /// </summary> 
    private readonly Action<System.IO.Stream> content; 

    private readonly bool bufferOutput; 

    /// <summary> 
    /// Initializes a new instance of the <see cref="FileGeneratingResult" /> class. 
    /// </summary> 
    /// <param name="fileName">Name of the file.</param> 
    /// <param name="contentType">Type of the content.</param> 
    /// <param name="content">Delegate with Stream parameter. This is the stream to which content should be written.</param> 
    /// <param name="bufferOutput">use output buffering. Set to false for large files to prevent OutOfMemoryException.</param> 
    public FileGeneratingResult(string fileName, string contentType, Action<System.IO.Stream> content,bool bufferOutput=true) 
     : base(contentType) 
    { 
     if (content == null) 
      throw new ArgumentNullException("content"); 

     this.content = content; 
     this.bufferOutput = bufferOutput; 
     FileDownloadName = fileName; 
    } 

    /// <summary> 
    /// Writes the file to the response. 
    /// </summary> 
    /// <param name="response">The response object.</param> 
    protected override void WriteFile(System.Web.HttpResponseBase response) 
    { 
     response.Buffer = bufferOutput; 
     content(response.OutputStream); 
    } 
} 

控制器方法现在将是这样的：

public ActionResult Export(int id) 
{ 
    return new FileGeneratingResult(id + ".csv", "text/csv", 
     stream => this.GenerateExportFile(id, stream)); 
} 

public void GenerateExportFile(int id, Stream stream) 
{ 
    stream.Write(/**/); 
} 

请注意，如果缓冲关闭，

stream.Write(/**/); 

变得非常缓慢。解决方案是使用BufferedStream。在一种情况下，性能提高了约100倍。见

Unbuffered Output Very Slow

Answer 5

如果你不想得到你自己的结果类型，你可以简单地写Response.OutputStream并返回new EmptyResult()。