How to pass default extension to rs.OpenFileName()?

Scripting
5

@dale, thanks. It seems that OpenFileDialog.DefaultExt serves a different purpose than i expected. So i was able to use FilterIndex instead in the example function below.

import System
import os

def OpenFileNameEx(title=None, filter=None, folder=None, filename=None, filter_index=None):

    ofd = System.Windows.Forms.OpenFileDialog()

    if title != None:
        if isinstance(title, str): ofd.Title = title

    if filter != None:
        if isinstance(title, str): ofd.Filter = filter

    if folder != None: 
        if os.path.isdir(folder): ofd.InitialDirectory = folder

    if filename != None:
        if isinstance(filename, str): ofd.FileName = filename

    if filter_index != None:
        if isinstance(filter_index, int): ofd.FilterIndex = filter_index

    if ofd.ShowDialog() == System.Windows.Forms.DialogResult.OK:
        return ofd.FileName

c.