Hi!

You shold be able to simply call img.Save( uStream, System.Drawing.Imaging.ImageFormat.Png);
where "uStream" is the stream returned by uFile.Open();

/johan

The upload function seems to be working but as soon as I try to do something besides uploading the stream from the file picker I get an "empty" image. Or maybe I don't understans your answer. Where does the uStream and uFile come in?

Hi,

uFile from your code sample in the first post, uStream would be what you called "s":

Heres how it could look like:

UnifiedFile uFile = directory.CreateFile(fileName + "." + fileExtension);

Stream s = uFile.Open(FileMode.CreateNew);

using (System.Drawing.Bitmap img = ResizeImage(new System.Drawing.Bitmap(profilePicture.PostedFile.InputStream), 151, 500, ResizeOptions.MaxWidth))
{
    img.Save(s, System.Drawing.Imaging.ImageFormat.Png);
}

Yeah, but that part of the code is fully functional and I wouldn't like to mess with it. It's how I call that function correctly that is the question. The "Stream fileContent" input parameter works great when I call it and asign it with fileUpload.PostedFile.InputStream but I first need to do something like this, but that gives me the faulty image:

System.Drawing.Bitmap img = ResizeImage(new System.Drawing.Bitmap(fileUpload.PostedFile.InputStream), 151, 500, ResizeOptions.MaxWidth);

MemoryStream ms = newMemoryStream();
img.Save(ms, ImageFormat.Png);
fileHandler.UploadFile(ms, HostingEnvironment.VirtualPathProvider.GetDirectory(path.VirtualPath + subDir) asUnifiedDirectory,  userID, "png", true);

As soon as I asign it with ms instead of fileUpload.PostedFile.InputStream something goes wrong. It's as if something is wrong with the ms-stream.

Ok, I see, you dont want to modify your UploadFile() function?

Try to simply reset (seek) the MemoryStream to its beginning before calling UploadFile().

img.Save(ms, ...);
ms.Seek( 0, SeekOrigin.Begin );
UploadFile(ms, ... );

But; in order to preserve server memory, it would be beneficial if the MemoryStream-step could be
avoided altogether and the (resized)img could save to the UnifiedFile's stream directly.

/jiohan

That worked, thanks!

Yes, you are right about the whole memory preserving but that would mean one function for each kind of upload, right? I was trying to keep the code to a minimun.

Not necessarily, there are numerous ways you can refactor that function, here's one way that contains two overloaded UploadFile() functions, one with the same arguments as before, and one new that lets you pass in a lambdaexpression for the code to write to the uniifiedFileStream.

        public string UploadFile(Stream fileContent, UnifiedDirectory directory, string fileName, string fileExtension, bool overwriteExisting)
        {
            return UploadFile(directory, fileName, fileExtension, overwriteExisting, (Stream writeStream) =>
            {
                int bytesRead = 0;
                int numBytesToRead = 2048;
                byte[] buffer = new byte[numBytesToRead];
                
                while(0 < (bytesRead = fileContent.Read(buffer, 0, numBytesToRead))
                {
                    writeStream.Write(buffer, 0, bytesRead);
                }    

                fileContent.Close();
            });
        }

        public string UploadFile(UnifiedDirectory directory, string fileName, string fileExtension, bool overwriteExisting, Action<Stream> writeStream)
        {
            UnifiedFile uFile = directory.CreateFile(fileName + "." + fileExtension);
            Stream s = uFile.Open(FileMode.CreateNew);

            writeStream(s);
            s.Close();

            return uFile.VirtualPath;
        }

With this modification, you could now call the UploadFile() function like this (when dealing with your resized bitmaps)

        UploadFile(Directory, fileName, fileExtension, overwriteExisting, (Stream s) => {

            img.Save( s, ImageFormat.Png );

        });

/johan

Wow, thanks! Appreciate the effort!

Thanks johan. Helped me alot.