How to get the file in byte[] in c# windows form using openfile dialog

Question

I want to retrieve mail attachments and store them into bytes[] so that I can insert them into table where a column attachment is of varbinary type. In windows form I am using OpenFileDialog and I don't know how to get the file's bytes in a byte[] type so that I can insert that into table in SQLServer.

Here is the code:

namespace fyp8
{
    public partial class ComposeMail : Form1
    {
        ArrayList alAttachments;
        MailMessage mmsg;
        string conn = @"Data source=(local);Initial Catalog=fyp;Integrated Security=true";


        public ComposeMail()
        {
            InitializeComponent();
        }

        private void ComposeMail_Load(object sender, EventArgs e)
        {

        }

        private void btnAttachment_Click(object sender, EventArgs e)
        {
            OpenFileDialog odflg = new OpenFileDialog();
            odflg.ShowDialog();
            if (odflg.ShowDialog() == DialogResult.OK)
            {

                try
                {
                    txtAttachments.Text = odflg.FileName;

                }

                catch(Exception ex)
                {
                    MessageBox.Show(ex.Message, "Error");
                }
            }
        }

        private void btnSend_Click(object sender, EventArgs e)
        {

            try
            {
                if (txtEmid.Text != null && txtPass.Text != null)
                {
                    mmsg = new MailMessage(txtEmid.Text, txtTo.Text);

                    mmsg.Subject = txtSub.Text;
                    mmsg.Body = txtBody.Text;
                    mmsg.IsBodyHtml = true;

                    /* Set the SMTP server and send the email with attachment */

                    SmtpClient smtpClient = new SmtpClient();

                    // smtpClient.Host = emailServerInfo.MailServerIP;
                    //this will be the host in case of gamil and it varies from the service provider

                    smtpClient.Host = "smtp.gmail.com";
                    //smtpClient.Port = Convert.ToInt32(emailServerInfo.MailServerPortNumber);
                    //this will be the port in case of gamil for dotnet and it varies from the service provider

                    smtpClient.Port = 587;
                    smtpClient.UseDefaultCredentials = true;

                    //smtpClient.Credentials = new System.Net.NetworkCredential(emailServerInfo.MailServerUserName, emailServerInfo.MailServerPassword);
                    smtpClient.Credentials = new NetworkCredential(txtEmid.Text, txtPass.Text);

                    //Attachment
                    Attachment attachment = new Attachment(txtAttachments.Text);
                    if (attachment != null)
                    {
                        mmsg.Attachments.Add(attachment);
                    }


                    //this will be the true in case of gamil and it varies from the service provider
                    smtpClient.EnableSsl = true;
                    smtpClient.Send(mmsg);
                }

                string msg = "Message Send Successfully:";
                msg += "\n To :" + txtTo.Text;

                SqlConnection con = new SqlConnection(conn);
                string query = "insert into sentmail values(@from,@to,@sub,@body,@status)";
                SqlCommand cmd = new SqlCommand(query, con);
                cmd.Parameters.AddWithValue("@from", txtFrom.Text);
                cmd.Parameters.AddWithValue("@to", txtTo.Text);
                cmd.Parameters.AddWithValue("@sub", txtSub.Text);
                cmd.Parameters.AddWithValue("@body",txtBody.Text);
                cmd.Parameters.AddWithValue("@status","sent" );
                con.Open();
                cmd.ExecuteNonQuery();
                con.Close();

                MessageBox.Show(msg.ToString());

                /* clear the controls */
                txtEmid.Text = string.Empty;
                txtPass.Text = string.Empty;
                txtFrom.Text = string.Empty;
                txtTo.Text = string.Empty;
                txtSub.Text = string.Empty;
                txtBody.Text = string.Empty;
                txtAttachments.Text = string.Empty;
            }

            catch (Exception ex)
            { 
                MessageBox.Show(ex.Message.ToString());
            }
        }
    }
}

Can you please provide a minimal example? From what I gather from your question text, all the code related to the open dialog and the e-mail is entirely irrelevant to your problem. You have a filename string, and you want to get the contents of that file into a byte array, right? — O. R. Mapper
– O. R. Mapper, Commented Apr 26, 2014 at 22:08

Jonathon Reinhart · Accepted Answer · 2014-04-26 22:10:45Z

14

Use the FileDialog.FileName property to get the selected file name.
Use File.ReadAllBytes to easily read the entire file into a byte array.

void Test() {
    var dialog = new OpenFileDialog();
    var result = dialog.ShowDialog();
    if (result != DialogResult.OK)
        return;

    byte[] buffer = File.ReadAllBytes(dialog.FileName);

    // Do whatever you want here with buffer
}

You're currently calling ShowDialog() twice. Just call it once and save the result.

answered Apr 26, 2014 at 22:10

Jonathon Reinhart

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1 Comment

user3469177 Over a year ago

when i declare var dialog=Openfiledialog() on the button click event its giving the erreo as system.windows.form.openfiledialog is a type but is used like variable.

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