I used information from here http://msdn.microsoft.com/ru-ru/library/system.windows.forms.openfiledialog(v=vs.110).aspx
this way:
Microsoft.Win32.OpenFileDialog dlg = new Microsoft.Win32.OpenFileDialog();
dlg.DefaultExt = ".xml"; // this is how I get only required extension
dlg.Filter = "XML files (*.xml)|*.xml"; // I guess, this should be modified, don't know how.
dlg.InitialDirectory = _directoryName1;
// here we go
Nullable<bool> result = dlg.ShowDialog();
if (result == true)
{
string path = dlg.FileName;
In the Initial Directory I have to types of files of same xml
extension, which names begin with script-Data...
or GeneralParam...
. So I need to show in the OpenFileDialog only files, which names begin with script-Data...
.
I know, that I can notify user, what he has decided wrong file by parsing path
, but it isn't good solution for me. Is any other way out here?
you have already set the Filter
property. so you can see only .xml
files when OpenFileDialog
opens. but if you want to filter
the filenames
to be displayed in OpenFileDialog
you can set the FileName
property as there is no other option to filter
by filename
Try This:
dlg.FileName = "script-Data*";