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]))]
>>
>>
>>
>>
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