Solved Windows Explorer opening location...

I found a tutorial that almost does what I want:
http://www.vistax64.com/tutorials/71306-windows-explorer-default-opening-3.html#post1048841
but it doesn't include the option I am looking for.

When I start Windows Explorer I want the folder pane to look like this:

Desktop
-myusername
-Public <--- this one open on start
-Computer
-Network
-Control Panel
-Recycle Bin

I know I can use "%SystemRoot%\explorer.exe c:\users\public" to access the same folder, but then the folder pane will look like this:

Desktop
-myusername
-Public
-Computer
--C:
---Users
----Public <--- this one open on start
-Network
-Control Panel
-Recycle Bin

I don't really want to start with the whole directory tree open. I assume there is a registry key-like thing for this? It would have to be similar syntax to starting with 'Computer' open.

Thanks

Dunno if this is what you want, but try:

"%SystemRoot%\explorer.exe %public%"

Nope. Same effect as "%SystemRoot%\explorer.exe c:\users\public" above.

This is why I'm thinking it has to be like opening to the Recycle Bin:
"%SystemRoot%\explorer.exe /N,::{645FF040-5081-101B-9F08-00AA002F954E}"

I just don't know how to figure out what hex code I need.

%SystemRoot%\explorer.exe /N,::{4336a54d-038b-4685-ab02-99bb52d3fb8b}

The above line does what I want. I found it in the registry under:

HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\explorer\FolderDescriptions\{DFDF76A2-C82A-4D63-906A-5644AC457385}

the Parsing Name data holds the necessary hex stuff.