470
votes

Can I create a Controller that simply returns an image asset?

I would like to route this logic through a controller, whenever a URL such as the following is requested:

www.mywebsite.com/resource/image/topbanner

The controller will look up topbanner.png and send that image directly back to the client.

I've seen examples of this where you have to create a View - I don't want to use a View. I want to do it all with just the Controller.

Is this possible?

19
I asked a similar question here https://stackguides.com/questions/155906/creating-a-private-photo-gallery-using-aspnet-mvc and ended up finding a great guide to do this. I created an ImageResult class following this guide. https://blog.maartenballiauw.be/post/2008/05/13/aspnet-mvc-custom-actionresult.htmlVyrotek
If you're wanting to modify the image, use the ImageResizing.Net HttpModule for the best performance. If you don't, a FilePathResult adds only a few percent of overhead. URL rewriting adds slightly less.Lilith River
Why not using WebApi Controller instead of MVC? ApiController classA-Sharabiani

19 Answers

544
votes

Use the base controllers File method.

public ActionResult Image(string id)
{
    var dir = Server.MapPath("/Images");
    var path = Path.Combine(dir, id + ".jpg"); //validate the path for security or use other means to generate the path.
    return base.File(path, "image/jpeg");
}

As a note, this seems to be fairly efficient. I did a test where I requested the image through the controller (http://localhost/MyController/Image/MyImage) and through the direct URL (http://localhost/Images/MyImage.jpg) and the results were:

  • MVC: 7.6 milliseconds per photo
  • Direct: 6.7 milliseconds per photo

Note: this is the average time of a request. The average was calculated by making thousands of requests on the local machine, so the totals should not include network latency or bandwidth issues.

132
votes

Using the release version of MVC, here is what I do:

[AcceptVerbs(HttpVerbs.Get)]
[OutputCache(CacheProfile = "CustomerImages")]
public FileResult Show(int customerId, string imageName)
{
    var path = string.Concat(ConfigData.ImagesDirectory, customerId, "\\", imageName);
    return new FileStreamResult(new FileStream(path, FileMode.Open), "image/jpeg");
}

I obviously have some application specific stuff in here regarding the path construction, but the returning of the FileStreamResult is nice and simple.

I did some performance testing in regards to this action against your everyday call to the image (bypassing the controller) and the difference between the averages was only about 3 milliseconds (controller avg was 68ms, non-controller was 65ms).

I had tried some of the other methods mentioned in answers here and the performance hit was much more dramatic... several of the solutions responses were as much as 6x the non-controller (other controllers avg 340ms, non-controller 65ms).

103
votes

To expland on Dyland's response slightly:

Three classes implement the FileResult class:

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

They're all fairly self explanatory:

  • 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.

Below is an example of using the content-disposition technique (not tested):

    [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;
    }
74
votes

This might be helpful if you'd like to modify the image before returning it:

public ActionResult GetModifiedImage()
{
    Image image = Image.FromFile(Path.Combine(Server.MapPath("/Content/images"), "image.png"));

    using (Graphics g = Graphics.FromImage(image))
    {
        // do something with the Graphics (eg. write "Hello World!")
        string text = "Hello World!";

        // Create font and brush.
        Font drawFont = new Font("Arial", 10);
        SolidBrush drawBrush = new SolidBrush(Color.Black);

        // Create point for upper-left corner of drawing.
        PointF stringPoint = new PointF(0, 0);

        g.DrawString(text, drawFont, drawBrush, stringPoint);
    }

    MemoryStream ms = new MemoryStream();

    image.Save(ms, System.Drawing.Imaging.ImageFormat.Png);

    return File(ms.ToArray(), "image/png");
}
12
votes

You can create your own extension and do this way.

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")%>
11
votes

Why not go simple and use the tilde ~ operator?

public FileResult TopBanner() {
  return File("~/Content/images/topbanner.png", "image/png");
}
10
votes

You can write directly to the response but then it isn't testable. It is preferred to return an ActionResult that has deferred execution. Here is my resusable StreamResult:

public class StreamResult : ViewResult
{
    public Stream Stream { get; set; }
    public string ContentType { get; set; }
    public string ETag { get; set; }

    public override void ExecuteResult(ControllerContext context)
    {
        context.HttpContext.Response.ContentType = ContentType;
        if (ETag != null) context.HttpContext.Response.AddHeader("ETag", ETag);
        const int size = 4096;
        byte[] bytes = new byte[size];
        int numBytes;
        while ((numBytes = Stream.Read(bytes, 0, size)) > 0)
            context.HttpContext.Response.OutputStream.Write(bytes, 0, numBytes);
    }
}
6
votes

UPDATE: There are better options than my original answer. This works outside of MVC quite well but it's better to stick with the built-in methods of returning image content. See up-voted answers.

You certainly can. Try out these steps:

  1. Load the image from disk in to a byte array
  2. cache the image in the case you expect more requests for the image and don't want the disk I/O (my sample doesn't cache it below)
  3. Change the mime type via the Response.ContentType
  4. Response.BinaryWrite out the image byte array

Here's some sample code:

string pathToFile = @"C:\Documents and Settings\some_path.jpg";
byte[] imageData = File.ReadAllBytes(pathToFile);
Response.ContentType = "image/jpg";
Response.BinaryWrite(imageData);

Hope that helps!

5
votes

Solution 1: To render an image in a view from an image URL

You can create your own extension method:

public static MvcHtmlString Image(this HtmlHelper helper,string imageUrl)
{
   string tag = "<img src='{0}'/>";
   tag = string.Format(tag,imageUrl);
   return MvcHtmlString.Create(tag);
}

Then use it like:

@Html.Image(@Model.ImagePath);

Solution 2: To render image from database

Create a controller method that returns image data like below

public sealed class ImageController : Controller
{
  public ActionResult View(string id)
  {
    var image = _images.LoadImage(id); //Pull image from the database.
    if (image == null) 
      return HttpNotFound();
    return File(image.Data, image.Mime);
  }
}

And use it in a view like:

@ { Html.RenderAction("View","Image",new {[email protected]})}

To use an image rendered from this actionresult in any HTML, use

<img src="http://something.com/image/view?id={imageid}>
5
votes

This worked for me. Since I'm storing images on a SQL Server database.

    [HttpGet("/image/{uuid}")]
    public IActionResult GetImageFile(string uuid) {
        ActionResult actionResult = new NotFoundResult();
        var fileImage = _db.ImageFiles.Find(uuid);
        if (fileImage != null) {
            actionResult = new FileContentResult(fileImage.Data,
                fileImage.ContentType);
        }
        return actionResult;
    }

In the snippet above _db.ImageFiles.Find(uuid) is searching for the image file record in the db (EF context). It returns a FileImage object which is just a custom class I made for the model and then uses it as FileContentResult.

public class FileImage {
   public string Uuid { get; set; }
   public byte[] Data { get; set; }
   public string ContentType { get; set; }
}
4
votes

you can use File to return a file like View, Content etc

 public ActionResult PrintDocInfo(string Attachment)
            {
                string test = Attachment;
                if (test != string.Empty || test != "" || test != null)
                {
                    string filename = Attachment.Split('\\').Last();
                    string filepath = Attachment;
                    byte[] filedata = System.IO.File.ReadAllBytes(Attachment);
                    string contentType = MimeMapping.GetMimeMapping(Attachment);

                    System.Net.Mime.ContentDisposition cd = new System.Net.Mime.ContentDisposition
                    {
                        FileName = filename,
                        Inline = true,
                    };

                    Response.AppendHeader("Content-Disposition", cd.ToString());

                    return File(filedata, contentType);          
                }
                else { return Content("<h3> Patient Clinical Document Not Uploaded</h3>"); }

            }
4
votes

Below code utilizes System.Drawing.Bitmap to load the image.

using System.Drawing;
using System.Drawing.Imaging;

public IActionResult Get()
{
    string filename = "Image/test.jpg";
    var bitmap = new Bitmap(filename);

    var ms = new System.IO.MemoryStream();
    bitmap.Save(ms, ImageFormat.Jpeg);
    ms.Position = 0;
    return new FileStreamResult(ms, "image/jpeg");
}
3
votes

Look at ContentResult. This returns a string, but can be used to make your own BinaryResult-like class.

2
votes
if (!System.IO.File.Exists(filePath))
    return SomeHelper.EmptyImageResult(); // preventing JSON GET/POST exception
else
    return new FilePathResult(filePath, contentType);

SomeHelper.EmptyImageResult() should return FileResult with existing image (1x1 transparent, for example).

This is easiest way if you have files stored on local drive. If files are byte[] or stream - then use FileContentResult or FileStreamResult as Dylan suggested.

1
votes

I see two options:

1) Implement your own IViewEngine and set the ViewEngine property of the Controller you are using to your ImageViewEngine in your desired "image" method.

2) Use a view :-). Just change the content type etc.

1
votes

You could use the HttpContext.Response and directly write the content to it (WriteFile() might work for you) and then return ContentResult from your action instead of ActionResult.

Disclaimer: I have not tried this, it's based on looking at the available APIs. :-)

0
votes

I also encountered similar requirement,

So in my case I make a request to Controller with the image folder path, which in return sends back a ImageResult object.

Following code snippet illustrate the work:

var src = string.Format("/GenericGrid.mvc/DocumentPreviewImageLink?fullpath={0}&routingId={1}&siteCode={2}", fullFilePath, metaInfo.RoutingId, da.SiteCode);

                if (enlarged)
                    result = "<a class='thumbnail' href='#thumb'>" +
                        "<img src='" + src + "' height='66px' border='0' />" +
                        "<span><img src='" + src + "' /></span>" +
                        "</a>";
                else
                    result = "<span><img src='" + src + "' height='150px' border='0' /></span>";

And in the Controller from the the image path I produce the image and return it back to the caller

try
{
  var file = new FileInfo(fullpath);
  if (!file.Exists)
     return string.Empty;


  var image = new WebImage(fullpath);
  return new ImageResult(new MemoryStream(image.GetBytes()), "image/jpg");


}
catch(Exception ex)
{
  return "File Error : "+ex.ToString();
}
0
votes

Read the image, convert it to byte[], then return a File() with a content type.

public ActionResult ImageResult(Image image, ImageFormat format, string contentType) {
  using (var stream = new MemoryStream())
    {
      image.Save(stream, format);
      return File(stream.ToArray(), contentType);
    }
  }
}

Here are the usings:

using System.Drawing;
using System.Drawing.Imaging;
using System.IO;
using Microsoft.AspNetCore.Mvc;
-1
votes

Yes you can return Image

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

(Please don't forget to mark this as answer)