You can do it:
db.tblvehicles.Model.represent = Lambda value,row: value.MakeID.Make

but unless you denormalize this is slow (two nested select per row).


On Tuesday, 7 January 2014 22:02:18 UTC-6, MVolpes wrote:
>
> Hi 
>
> I have been using this framework for the last few days and I am in awe a 
> fantastic framework. I have come from a strong .net and php framework back 
> ground and wanted to learn python this has already taught me so much!!
>
> I have a question that hopefully can be answered easily and maybe I am 
> overlooking something.
>
> I have the smart grid working well however i want to display a reference 
> of a reference eg
>
> Vehicle > Model > Make
>
>
>
> Please bare in mind i am using a legacy db
>
> Model
> #--------
> db.define_table('tblmake',
>     Field('MakeID','id'),
>     Field('Make','string'),
>     format="%(Make)s",
>     migrate=False)
>
>
> #--------
> db.define_table('tblmodels',
>     Field('Model','string'),
>     Field('ModelID','id'),
>     Field('MakeID','reference tblmake'),
>     format="%(Model)s",
>     migrate=False)
>
>
> #--------
> db.define_table('tblvehicles',
>     Field('VehicleID','id'),
>     Field('Registration','string'),
>     Field('Driver','string'),
>     Field('Model','reference tblmodels'),
>     migrate=False)
>
> Controller
>
> vehiclefields = [db.tblvehicles.Registration, db.tblvehicles.Driver, 
> db.tblvehicles.Model]
>
> vehiclelinks = [lambda row: A('View 
> Post',_href=URL("search","test",args=[row.VehicleID]))]
> vehicles = SQLFORM.smartgrid(db.tblvehicles, orderby="VehicleID Desc", 
> fields=vehiclefields, formname="vehiclegrid",
>                                  details=False, editable=False, 
> deletable=False,links=vehiclelinks, linked_tables=False)
>
> I would like to display the db.tblmake.Make field 
>
> How can I do this?
> Can I do this?
>
> I have tried adding various combinations but with no luck
>
> Any help would be greatly appreciated
>
>

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