Set path for file open command programmatically

Question

I am working on a python addin for ArcGIS 10.2.2 and I would like to set the default path for the "file open" command to a user-specified location after a script tool executes.  One of the functions that my addin performs is to launch a script tool that copies some .mxd's to a user-specified location.  I would like to use that location as the default path, so users can access the other mxd's that were copied by the tool without having to navigate to that folder.  I have used the comtypes module to get a reference to the current document, and it seems I can wrap any of the ArcObjects interfaces I will need.  Is there a way to access the property that the Open command uses to open the explorer window?  Any suggestions about where to start digging into ArcObjects would be appreciated.

OwainCatton · Answer

Glad you found it alright, my mistake giving you the Arc Catalog registry key location.

Owain

MarshallUdo · Answer

The entry that controls the path is here:

HKEY_CURRENT_USER\Software\ESRI\Desktop10.2\ArcMap\File Path

Thanks to Owain for providing the example on how to change the value for this key using python.

MarshallUdo · Answer

Thank you for your reply Owain.  I looked in the registry, and on my machine (Windows Server 2008 R2 with ArcGIS 10.2.2), the LastBrowseLocation entry is here:

HKEY_CURRENT_USER\Software\ESRI\Desktop10.2\ArcCatalog\Settings\LastBrowseLocation

Unfortunately, the string doesn't match the path that opens in the open dialog.  I tried manually changing the value, and the open file dialog didn't open to the path that I specified.