How to open new OpenFileDialog automatically in Vi

Question

I'm on Vista and I'm using Microsoft.Win32.OpenFileDialog class.

When I call ShowDialog() I get the old XP-style dialog:

How do I get the new Vista-style dialog with fallback to the old one on WindowsXP?

A bit of rumble:

I don't really understand why they didn't replace the dialog in vista, but kept both of them. Now legacy apps will never open new dialog, unless updated.

Answer 1

The first dialog you showed is a save dialog not an open dialog.

You should only have to do this:

OpenFileDialog OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.InitialDirectory = "c:\\";
openFileDialog1.Filter = "My files (*.myfile)|*.myfile|All files (*.*)|*.*";
openFileDialog1.FilterIndex = 1;
openFileDialog1.RestoreDirectory = true;

if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
  //openFileDialog1.FileName
}

Answer 2

Reference System.Windows.Forms

using System.Windows.Forms

OpenFileDialog openFileDialog1 = new OpenFileDialog();

openFileDialog1.InitialDirectory = "c:\\" ;
openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*" ;
openFileDialog1.FilterIndex = 2 ;
openFileDialog1.RestoreDirectory = true ;

if(openFileDialog1.ShowDialog() == DialogResult.OK)
{
    //Do Stuff
}

Answer 3

Yes, you'd have to upgrade to .NET 4.0 to get the new dialog. If you're stuck on 3.5 then you can use System.Windows.Forms.OpenFileDialog, it did get the update to use the new IFileDialog COM interface.

The fallback is automatic but you can use its AutoUpgradeEnabled property to force legacy, if necessary. Which it is not, unlikely that a .NET program would modify the dialog.