Question

im totally stuck here on creating a random number. Ideally not the same number twice as its a customer number.

I then need to display it in my tb_id (textbox) and write the line into a customer file.

Any help here would be great.

private void button1_Click(object sender, EventArgs e)
{    
    **var rng = new Random();**
    // Here I need to write the random number (it's a customer number) to the tb_id text box. so I'm able to write it to their customer file afterwards.

    try
    {
        string fileName = string.Format(tb_surname.Text);

        if (tb_firstname.Text == "" || fileName == "" || tb_postcode.Text == "")
        {
            MessageBox.Show("Missing values from textboxes!");
        }
        else if (File.Exists(fileName + "Text"))
        {
            if (MessageBox.Show("Warning: There is already a file with the surname you enter already on the system. Contents will be replaced - do you want to continue?", "File Demo", MessageBoxButtons.YesNo, MessageBoxIcon.Question,
                 MessageBoxDefaultButton.Button1, MessageBoxOptions.DefaultDesktopOnly) == DialogResult.Yes)
            {
                //write lines of text to file
                StreamWriter outputStream = File.CreateText(fileName + ".Txt");
                outputStream.WriteLine(tb_firstname.Text);
                outputStream.WriteLine(tb_surname.Text);
                outputStream.WriteLine(tb_surname.Text);
                **outputStream.WriteLine(tb_id.Text);**
                outputStream.Close();
                MessageBox.Show("Text saved to file successfully!");
                this.Close();
            }
        }
    }
    // ...
}

part 2

       // creating a random number here for the customer id
        Guid g = Guid.NewGuid();
        tb_id.Text = g.ToString();

to write Guid number to a text file.

        outputStream.WriteLine(tb_id.Text);

Answer 1

I would not recomend using the random number if you want to get it unique for users there is a high probability of getting different users with same number.

guid is probably better idea

  Guid g = Guid.NewGuid();