nevermind I fixed it. Not sure why, but if it was named 'equipment' the controller showed up...any other name it doesnt. I just renamed
On Monday, August 27, 2012 5:05:44 PM UTC-6, SeamusSeamus wrote: > > ah, yes, all is well now...however, this is strange. > Now, when I click the link on the table to take me to the > 'equipment.html' page (I renamed it from details.html), it shows the > controller in the URL: > www.mysite.com/default/equipment/id/title > > how do I make it so the 'default' does not show up? It didnt show up > before it looked like this: > > links = [lambda row: A('Details',_href=URL('default', 'details', args=[ > row.id, row.slug]))] > I changed to > links = [lambda row: A('Details',_href=URL('default', 'equipment', > args=[row.id, row.slug]))] > > I also changed the controller from def default to def equipment, and also > changed default.html to equipment.html in views... > > now it shows the controller in the URL... > > my routes looks like this > > routers = dict( > BASE = dict(default_application='equipment', > default_controller = 'default',), > ) > > > On Monday, August 27, 2012 4:38:41 PM UTC-6, Limedrop wrote: >> >> >> I don't know if you picked this up, but you don't seem to have row.id as >> part of the sqlfrom url. >> >> Perhaps try: >> links = [lambda row: A('Details',_href=URL('default','equipment', >> args=[row.id, row.slug]))] >> >> >> On Tuesday, August 28, 2012 10:28:39 AM UTC+12, SeamusSeamus wrote: >>> >>> Okay, but do I leave everything else alone? All I want to do is make it >>> www.mysite.com/equipment/id/title >>> currently, title is set up as slug in the DB. What am I doing wrong. >>> >>> In my view: >>> >>> equipment_id = request.args(0) >>> equipment_slug = request.args(1) >>> query = (db.equipment.id == item_id) & (db.equipment.slug == >>> item_slug) >>> item = db(query).select().first() >>> >>> try: >>> equipment = db.equipment[int(request.args(0))] >>> except: >>> equipment = db.equipment(request.args(0)) or >>> db(db.equipment.slug == request.args(0)).select().first() >>> >>> >>> and then my link in sqlform: >>> >>> links = [lambda row: A('Details',_href=URL('default','equipment', >>> args=[row.slug]))] >>> >>> >>> >>> --