How can I extract parts of the EXIF metadata to use in a Smart Collection?

Question

I want to aggregate all of the images I've captured by Capture Month and identify those images that were scanned or have no Capture Date.  This question actually attempts to open the door for read access to all of the image metadata and the ability to store some of it in user-identified fields which, somehow, the user will be able to populate.
So - if the date fields (i.e. DateTime Original, DateTime Captured, DateTime Modified ) were broken down into, for example using DateTime Original:  Original Year-YYYY, Original Month-MM, Original Day-DD.  It may be possible to do this with a redefination of the schema used to define each of the fields.  That would be the simpliest, easiest way - if that's possible.
If it isn't possible (or not in the near future, anyway) does anyone have any ideas?
Thanks

Answer 1

With over 45 years in IT (I know, it only means I'm an old man - not that I learned anything ), I observed that many of the problems I identified had simple solutions.  The reason they weren't in the original releases was, the testers felt "Why would someone want to do that?  Or need that data?  Or that field?"  Why would you hold an iPhone quite like THAT?
Since Adobe defined the schema that captures the image metadata and defined what fields would be accessable and in what way; the most direct solution to this is to modify the schema through a redefinition of the date fields — all of them — so the individual elements are accessable.
Also, because one of Lightroom's key features is image management, these new data fields should be accessible for Smart Collections.  There's at least one other person in this community who's interested in a By Month Across Year sort and I suspect many people don't speak up, even if they have a similar need.
That's what I meant.  While outside developers (non-Adobe employees) might be able to come up with a patch for the solution, the underlying problem will still remain until Adobe acts.  Solving the problem is the less expensive means of dealing with the issue.
I think.