Getting an Image object from a byte array

Question

I\'ve got a byte array for an image (stored in the database). I want to create an Image object, create several Images of different sizes and store them back in the database (sa

Answer 1

Based on your comments to another answer, you can try this for performing a transformation on an image that's stored in a byte[] then returning the result as another byte[].

public byte[] TransformImage(byte[] imageData)
{
    using(var input = new MemoryStream(imageData))
    {
        using(Image img = Image.FromStream(input))
        {
            // perform your transformations

            using(var output = new MemoryStream())
            {
                img.Save(output, ImageFormat.Bmp);

                return output.ToArray();
            }
        }
    }
}

This will allow you to pass in the byte[] stored in the database, perform whatever transformations you need to, then return a new byte[] that can be stored back in the database.

Answer 2

I thought I'd add this as an answer to make it more visible.

With saving it back to a byte array:

    public Image localImage;
    public byte[] ImageBytes;

    public FUU_Image(byte[] bytes)
    {
        using (MemoryStream ImageStream = new System.IO.MemoryStream(bytes))
        {
            localImage = Image.FromStream(ImageStream);
        }

        localImage = ResizeImage(localImage);

        using (MemoryStream ImageStreamOut = new MemoryStream())
        {
            localImage.Save(ImageStreamOut, ImageFormat.Jpeg);
            ImageBytes = ImageStreamOut.ToArray();
        }

    }

Answer 3

You'd use a MemoryStream to do this:

byte[] bytes;
...
using (var ms = new System.IO.MemoryStream(bytes)) {
   using(var img = Image.FromStream(ms)) {
      ...
   }
}

Answer 4

Only answering the first half of the question: Here's a one-liner solution that works fine for me with a byte array that contains an image of a JPEG file.

Image x = (Bitmap)((new ImageConverter()).ConvertFrom(jpegByteArray));

EDIT: And here's a slightly more advanced solution: https://stackoverflow.com/a/16576471/253938