是的，你可以返回图像

public ActionResult GetImage(string imageFileName)
{
    var path = Path.Combine(Server.MapPath("/Images"), imageFileName + ".jpg"); 
    return base.File(path, "image/jpeg");
}

(请不要忘记将此标记为答案)

我有两个选择:

1)实现你自己的IViewEngine，设置控制器的ViewEngine属性，你正在使用你的ImageViewEngine在你想要的“image”方法。

2)使用视图:-)。只需改变内容类型等。

查看ContentResult。这将返回一个字符串，但可用于创建您自己的类binaryresult。

您可以创建自己的扩展，并这样做。

public static class ImageResultHelper
{
    public static string Image<T>(this HtmlHelper helper, Expression<Action<T>> action, int width, int height)
            where T : Controller
    {
        return ImageResultHelper.Image<T>(helper, action, width, height, "");
    }

    public static string Image<T>(this HtmlHelper helper, Expression<Action<T>> action, int width, int height, string alt)
            where T : Controller
    {
        var expression = action.Body as MethodCallExpression;
        string actionMethodName = string.Empty;
        if (expression != null)
        {
            actionMethodName = expression.Method.Name;
        }
        string url = new UrlHelper(helper.ViewContext.RequestContext, helper.RouteCollection).Action(actionMethodName, typeof(T).Name.Remove(typeof(T).Name.IndexOf("Controller"))).ToString();         
        //string url = LinkBuilder.BuildUrlFromExpression<T>(helper.ViewContext.RequestContext, helper.RouteCollection, action);
        return string.Format("<img src=\"{0}\" width=\"{1}\" height=\"{2}\" alt=\"{3}\" />", url, width, height, alt);
    }
}

public class ImageResult : ActionResult
{
    public ImageResult() { }

    public Image Image { get; set; }
    public ImageFormat ImageFormat { get; set; }

    public override void ExecuteResult(ControllerContext context)
    {
        // verify properties 
        if (Image == null)
        {
            throw new ArgumentNullException("Image");
        }
        if (ImageFormat == null)
        {
            throw new ArgumentNullException("ImageFormat");
        }

        // output 
        context.HttpContext.Response.Clear();
        context.HttpContext.Response.ContentType = GetMimeType(ImageFormat);
        Image.Save(context.HttpContext.Response.OutputStream, ImageFormat);
    }

    private static string GetMimeType(ImageFormat imageFormat)
    {
        ImageCodecInfo[] codecs = ImageCodecInfo.GetImageEncoders();
        return codecs.First(codec => codec.FormatID == imageFormat.Guid).MimeType;
    }
}
public ActionResult Index()
    {
  return new ImageResult { Image = image, ImageFormat = ImageFormat.Jpeg };
    }
    <%=Html.Image<CapchaController>(c => c.Index(), 120, 30, "Current time")%>

if (!System.IO.File.Exists(filePath))
    return SomeHelper.EmptyImageResult(); // preventing JSON GET/POST exception
else
    return new FilePathResult(filePath, contentType);

SomeHelper.EmptyImageResult()应该返回具有现有图像的FileResult(例如1x1透明)。

这是最简单的方法，如果你有文件存储在本地驱动器。 如果文件是字节[]或流-然后使用FileContentResult或FileStreamResult Dylan建议。

稍微解释一下迪兰的回应:

有三个类实现了FileResult类:

System.Web.Mvc.FileResult
      System.Web.Mvc.FileContentResult
      System.Web.Mvc.FilePathResult
      System.Web.Mvc.FileStreamResult

它们都是不言自明的:

For file path downloads where the file exists on disk, use FilePathResult - this is the easiest way and avoids you having to use Streams. For byte[] arrays (akin to Response.BinaryWrite), use FileContentResult. For byte[] arrays where you want the file to download (content-disposition: attachment), use FileStreamResult in a similar way to below, but with a MemoryStream and using GetBuffer(). For Streams use FileStreamResult. It's called a FileStreamResult but it takes a Stream so I'd guess it works with a MemoryStream.

下面是一个使用内容处理技术的例子(未测试):

    [AcceptVerbs(HttpVerbs.Post)]
    public ActionResult GetFile()
    {
        // No need to dispose the stream, MVC does it for you
        string path = Path.Combine(AppDomain.CurrentDomain.BaseDirectory, "App_Data", "myimage.png");
        FileStream stream = new FileStream(path, FileMode.Open);
        FileStreamResult result = new FileStreamResult(stream, "image/png");
        result.FileDownloadName = "image.png";
        return result;
    }