One of the problems with using the OpenFileDialog class is that the FileName property always returns the full path name of the file:
OpenFileDialog openFileDialog1 = new OpenFileDialog()
{
    Filter = "Text files (*.jpg)*.jpg"
};
Console.WriteLine(openFileDialog1.FileName);  //---e.g. C:\Windows\text.txt---
To retrieve only the filename and not the path, feed it to the FileInfo class, like this:
FileInfo fi = new FileInfo(openFileDialog1.FileName);
Console.WriteLine(fi.Name);  //---text.txt---
Console.WriteLine(fi.Directory);  //---C:\Windows---
Cool, isn't it?
 
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