How to communicate OpenFileDialog result “FilePath” from view to view Model?

Question

I am trying to open a file dialog by pressing a button on view and view call a command "OpenFileDialog" which is defined in viewmodel. That method send a message to view to open the "Microsoft.Win32.OpenFileDialog()" and now I want to communicate the result of OpenFileDialog to ViewModel.

ViewModel

 public ICommand OpenFileCommand {
  get { return new RelayCommand( ( ) => OpenFileCommandExecute( ), ( ) => true ); }
}

private void OpenFileCommandExecute( ) {
  Messenger.Default.Send( "OpenfileDialog" );
}

View

 Messenger.Default.Register( this, "OpenFileDialog", openFileDialog) ;
 private void openFileDialog(  ) {
  OpenFileDialog OFP = new OpenFileDialog( );
  var kk = OFP.ShowDialog( );      
 }

I want to communicate the selected file path to view model. I am using MVVM Light Toolkit and WPF. My code is not in working condition.

sepai sepai · Accepted Answer · 2013-05-01T01:27:55

The OpenFileDialog should have a Filename property that you can check after the ShowDialog() method.

Example

OpenFileDialog openDialog = new OpenFileDialog();
openDialog.ShowDialog();
if (string.IsNullOrEmpty(openDialog.Filename)) return;

string path = openDialog.Filename;

Then if you want it accessible on the view model just create your own property for the path in the ViewModel class and populate it with the Filename property.

How to communicate OpenFileDialog result “FilePath” from view to view Model?

1 Answers