It should work. What did it output instead? Regards, Jason
On Sat, 2009-09-19 at 01:35 -0700, annet wrote: > Unfortunately it did not work. Every 'not so detailed item' gets its > own table instead of a table row. > > The latest function reads like: > > def latest(): > response.functionname='Laatste nieuws' > response.image=URL(r=request,c='static',f='init/media/ > utilities_1.jpg') > latest=db((db.nieuws.bedrijf==117)& > (db.nieuws.publicatie_datum<=request.now)&\ > (db.nieuws.publicatie_datum>=minus_three_months(request.now))) > .select(db.nieuws.ALL,orderby=~db.nieuws.publicatie_datum) > return dict(latest=latest) > > > Isn't it possible to divide latest into two dictionaries, one with the > first 6 items and the other with the next 18 items. > > def latest(): > response.functionname='Laatste nieuws' > response.image=URL(r=request,c='static',f='init/media/ > utilities_1.jpg') > latest6=db((db.nieuws.bedrijf==117)& > (db.nieuws.publicatie_datum<=request.now)&\ > (db.nieuws.publicatie_datum>=minus_three_months(request.now))) > .select(db.nieuws.ALL,orderby=~db.nieuws.publicatie_datum,limitby= > (0,5)) > latest18=db((db.nieuws.bedrijf==117)& > (db.nieuws.publicatie_datum<=request.now)&\ > (db.nieuws.publicatie_datum>=minus_three_months(request.now))) > .select(db.nieuws.ALL,orderby=~db.nieuws.publicatie_datum,limitby= > (6,23)) > return dict(latest6=latest6,latest18=latest18) > > > Kind regards, > > Annet. > > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "web2py-users" group. To post to this group, send email to web2py@googlegroups.com To unsubscribe from this group, send email to web2py+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/web2py?hl=en -~----------~----~----~----~------~----~------~--~---
