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]))]
>
>
>
>
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