How to Get Only the Name of a File

[Behseini]

Hi Guys,
I am trying to display a selected file name by user(using OpenFileDialog class and FileName property) as

if (dlgOpen.ShowDialog() == DialogResult.OK)
    {
     txtFileName.Text = dlgOpen.FileName;
    }

but the problem is this return all path info which I do not need! Can you please let me know How I can get rid of the path and just retrieve or display the file name only

Thanks for your time, in advanced.

[bhagawatshinde]

Try this...

if (dlgOpen.ShowDialog() == DialogResult.OK)

    {

     txtFileName.Text = dlgOpen.SafeFilename;

    }

[thines01]

...or Path.GetFileName();

using System;
using System.IO;

namespace DW_398724
{
   class Program
   {
      static void Main(string[] args)
      {
         string strFilepath = "c:/documents and settings/user/directory/sub_director/text.txt";
         Console.Write(Path.GetFileName(strFilepath)); // prints text.txt
      }
   }
}

[Behseini]

Thank you both bhagawatshinde and thines01,
it works now