C#

Quote:

The FileContent property gets a Stream object that points to a file to upload. Use this property to access the contents of the file as bytes. For example, you can use the Stream object that is returned by the FileContent property to read the contents of the file as bytes and store them in a byte array. Alternatively, you can use the FileBytes property to retrieve all the bytes in the file.

/// <summary>
/// Save an upload into the database.
/// </summary>
/// <param name="fl">Control containing the download.</param>
/// <returns>Status as a string</returns>
private string SaveUpload(FileUpload fl)
    {
    if (fl.HasFile)
        {
        try
            {
            int version = 0;
            string filename = Path.GetFileName(fl.FileName);
            byte[] filedata = fl.FileBytes;
            string strCon = System.Web.Configuration.WebConfigurationManager.ConnectionStrings["DownloadDatabase"].ConnectionString;
            using (SqlConnection con = new SqlConnection(strCon))
                {
                con.Open();
                // Check version - if the file exists in the DB already, then increase version number.
                using (SqlCommand ver = new SqlCommand("SELECT MAX(version) FROM dlContent WHERE fileName=@FN", con))
                    {
                    ver.Parameters.AddWithValue("@FN", filename);
                    object o = ver.ExecuteScalar();
                    if (o != null && o != System.DBNull.Value)
                        {
                        // Exists already.
                        version = (int) o + 1;
                        }
                    }
                // Stick file into database.
                using (SqlCommand ins = new SqlCommand("INSERT INTO dlContent (iD, fileName, description, dataContent, version, uploadedOn) " +
                                                       "VALUES (@ID, @FN, @DS, @DT, @VS, @UD)", con))
                    {
                    ins.Parameters.AddWithValue("@ID", Guid.NewGuid());
                    ins.Parameters.AddWithValue("@FN", filename);
                    ins.Parameters.AddWithValue("@DS", "");
                    ins.Parameters.AddWithValue("@DT", filedata);
                    ins.Parameters.AddWithValue("@VS", version);
                    ins.Parameters.AddWithValue("@UD", DateTime.Now);
                    ins.ExecuteNonQuery();
                    }
                }
            return string.Format("{0} uploaded, version = {1}", filename, version);
            }
        catch (Exception ex)
            {
            return "The file could not be uploaded. The following error occured: " + ex.Message;
            }
        }
    return "Please select a file.";
    }

Pete O'Hanlon wrote:

I just use a guid as the file name on the disk

Richard MacCutchan wrote:

No, it's because both my eyes and brain are getting more feeble

jrahma wrote:

but watermark now showing with no error

jrahma wrote:

My watermark is not showing error AND not showing
the watermark either!

Wes Aday wrote:

The code that you have posted works perfectly fine on my system.

Richard Andrew x64 wrote:

Works on My Machine

Quote:

To use this API, you must provide a manifest specifying Comclt32.dll version 6.0. For more information on manifests, see Enabling Visual Styles.

public static Tuple<double, double> Test(Double Val1, Double Val2, Double Val3)
{
    Double Test_Val = Val1 + Val2 + Val3;
    Double Test_2 = 7;
    return new Tuple<double, double>(Test_Val, Test_2);
}

static void Main(string[] args)
{
    Double A = 1, B = 2, C = 3;
    Tuple<double,double> Result = Test(A, B, C);
    Console.WriteLine("{0} {1}", Result.Item1 , Result.Item2 );
    Console.ReadLine();
}