I want to access/open file(s) that's why I used the openfiledialog.
This is my code:
that code is right, but I want to open a file that I choose from the OpenFileDialog1.
Whenever I choose a file and click the button Open form the dialogbox, nothing happens.
Can anyone knows how I will open the file that I choose from the dialog box??
thanks guys!
annir
This is my code:
PHP:
OpenFileDialog1.InitialDirectory = "//10.104.160.105"
OpenFileDialog1.ShowDialog()
If OpenFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK Then
OpenFileDialog1.OpenFile()
End If
that code is right, but I want to open a file that I choose from the OpenFileDialog1.
Whenever I choose a file and click the button Open form the dialogbox, nothing happens.
Can anyone knows how I will open the file that I choose from the dialog box??
thanks guys!
annir