Yes, the "max" solution was originally for a different problem that only needed the single max value and involved only one query. In this case, the orderby/limitby solution is the way to go.
Anthony On Wednesday, September 12, 2012 2:36:37 PM UTC-4, Niphlod wrote: > > ps: methods are NOT equivalent. They are if you have "continous" ids. > But, e.g., you remove some rows. You end up with > 1,2,3,4,5,6,7,8,9,10,11,13,15,17,20. > Second method (i.e. calc max and go back by ten) leaves you with > 20,17,15,13,11,10 (and takes two queries) > First method (i.e. orderby + limitby) correctly returns > 20,17,15,13,11,10,9,8,7,6. > > On Wednesday, September 12, 2012 7:09:39 PM UTC+2, Anthony wrote: >> >> On Wednesday, September 12, 2012 12:35:41 PM UTC-4, Massimo Di Pierro >> wrote: >>> >>> Instead of this >>> >>> maxID=db(db.node).select(db.node.id.max()).first()['MAX(node.id)'] >>> >>> I would do >>> >>> maxID=db(db.node).select(db.node.id.max()).first()[db.node.id.max()] >>> >> >> Well, that's what I really recommended: >> https://groups.google.com/d/msg/web2py/VY71mF2cl-4/oNE00OduXQ0J. :-) >> >> > --
