從動作寫入輸出流

Question

由於一些奇怪的原因，我想直接從控制器操作將HTML寫入Response流。 （我理解MVC分離，但這是一個特殊情況。）

我可以直接寫入HttpResponse流嗎？在那種情況下，控制器動作應該返回哪個IView對象？我可以返回'空'嗎？

Answer 1

你可以做return Content(...);的地方，如果我沒有記錯，...將要直接寫入到輸出流，或者什麼都沒有的東西。

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，你可以直接寫入響應。完成後，您可以調用CompleteRequest（），並且不需要返回任何內容。

例如：

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

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

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

Answer 3

寫自己的操作結果。這裏的礦井中的一個的示例：

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

我用從FileResult派生的類來實現這一點使用正常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); 
    } 
} 

控制器方法現在將是這樣的：

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(/**/); 
} 

請注意，如果緩衝關閉，

stream.Write(/**/); 

變得非常緩慢。解決方案是使用BufferedStream。在一種情況下，性能提高了約100倍。見

Unbuffered Output Very Slow

Answer 5

如果你不想得到你自己的結果類型，你可以簡單地寫Response.OutputStream並返回new EmptyResult()。