Open Windows Explorer to a Combined-GUID folder

Question

I want, for example, to open the "Font Settings" page. (It's under "Fonts" in the control panel.)

I got its GUID as follows:

::{26EE0668-A00A-44D7-9371-BEB064C98683}\1\Fonts\::{93412589-74D4-4E4E-AD0E-E0CB621440FD}

_{I got it using the URL of the NavigateComplete2 event of an InternetExplorer interface.}

I tried many different combinations such as:

Process.Start("explorer", @"::{26EE0668-A00A-44D7-9371-BEB064C98683}\1\Fonts\::{93412589-74D4-4E4E-AD0E-E0CB621440FD}");

And:

Process.Start("::{26EE0668-A00A-44D7-9371-BEB064C98683}", @"\1\Fonts\::{93412589-74D4-4E4E-AD0E-E0CB621440FD}");

But nothing works.

Answer 1

Painfully late answer, but your problem is/was that the url you were using is for some reason not accepted by the shell url handler.

To start that folder you want you can use: shell:::{26EE0668-A00A-44D7-9371-BEB064C98683}\1\::{93412589-74D4-4E4E-AD0E-E0CB621440FD}. If you really want the font folder to be the parent, try shell:Fonts\::{93412589-74D4-4E4E-AD0E-E0CB621440FD}. It gives you the absurd looking path "C:\Windows\Fonts\Font settings"