ASP.NET

// button click

byte[] img = new byte[FileUpload1.PostedFile.ContentLength];
            HttpPostedFile postfile = FileUpload1.PostedFile;
            postfile.InputStream.Read(img, 0, (int)FileUpload1.PostedFile.ContentLength);
            System.Drawing.Image img12 = System.Drawing.Image.FromStream(FileUpload1.PostedFile.InputStream);
            SqlCommand cmd = new SqlCommand("insert into image11 (type,name,length,byte) values(@type,@name,@length,@byte)", cnn);
            cmd.Parameters.Add("@type", SqlDbType.VarChar).Value = FileUpload1.PostedFile.ContentType;
            cmd.Parameters.Add("@name", SqlDbType.VarChar).Value = FileUpload1.PostedFile.FileName;
            cmd.Parameters.Add("@length", SqlDbType.VarChar).Value = FileUpload1.PostedFile.ContentLength;
            cmd.Parameters.Add("@byte", SqlDbType.Image, img.Length).Value = img;
            cnn.Open();
            cmd.ExecuteNonQuery();
            cnn.Close();
            DataTable dtTemp = new DataTable("TempImage");//Creating temprory data table which will store image information
            dtTemp.Columns.Add("Image", System.Type.GetType("System.Byte[]"));//Byte Image column
            DataRow dr = dtTemp.NewRow();
            dr[0] = img;//storing binary image information in table.
            dtTemp.Rows.Add(dr);
            Session["dt"] = dtTemp;//storing temprory table in seesion.
            Image1.ImageUrl = "de.aspx?id=0";