“Изображение в базу данных SQL” Ответ

Изображение в базу данных SQL

protected void btnUpload_Click(object sender, EventArgs e)
{
 string strImageName = txtName.Text.ToString();
 if (FileUpload1.PostedFile != null && 
     FileUpload1.PostedFile.FileName != "")
  {
   byte[] imageSize = new byte
                 [FileUpload1.PostedFile.ContentLength];
  HttpPostedFile uploadedImage = FileUpload1.PostedFile;
  uploadedImage.InputStream.Read
     (imageSize, 0, (int)FileUpload1.PostedFile.ContentLength);
 
 // Create SQL Connection 
  SqlConnection con = new SqlConnection();
  con.ConnectionString = ConfigurationManager.ConnectionStrings
                         ["ConnectionString"].ConnectionString;
 
 // Create SQL Command 
 
 SqlCommand cmd = new SqlCommand();
 cmd.CommandText = "INSERT INTO Images(ImageName,Image)" +
                   " VALUES (@ImageName,@Image)";
 cmd.CommandType = CommandType.Text;
 cmd.Connection = con;
 
 SqlParameter ImageName = new SqlParameter
                     ("@ImageName", SqlDbType.VarChar, 50);
 ImageName.Value = strImageName.ToString();
 cmd.Parameters.Add(ImageName);
 
 SqlParameter UploadedImage = new SqlParameter
               ("@Image", SqlDbType.Image, imageSize.Length);
 UploadedImage.Value = imageSize;
 cmd.Parameters.Add(UploadedImage);
 con.Open();
 int result = cmd.ExecuteNonQuery();
 con.Close();
 if (result > 0)
 lblMessage.Text = "File Uploaded";
 GridView1.DataBind();
 }
}