Hi, On Wed, Apr 4, 2012 at 9:06 AM, Thomas Mueller <[email protected]> wrote: > For Oak, COUNT(*) is definitely possible. Also important is GROUP BY I > believe, in combination with COUNT(*). I'm not sure about DISTINCT. > > But for Jackrabbit 2.x, I can't tell how easy it would be to support it.
I looked at this last year. It's doable if someone has the cycles for it, but requires quite a bit of effort as the relevant bits in Jackrabbit 2.x weren't designed with aggregate queries in mind. Implementing option 2 as outlined by Christian is far easier for this specific use case. BR, Jukka Zitting
