On Mon, Aug 06, 2007 at 03:24:28PM -0500, Jim Steil wrote:
> class RailCategory(SQLObject):
> class sqlmeta:
> style = Style(longID=True)
> idName = 'railCategoryId'
>
> name = UnicodeCol(length=50)
> description = UnicodeCol()
> railCars = RelatedJoin("RailCar", intermediateTable="railCategoryCar",
> joinColumn="railCategoryId",
> otherColumn="railCarId")
>
> class RailCar(SQLObject):
> class sqlmeta:
> style = Style(longID=True)
> idName = 'railCarId'
>
> carInitials = UnicodeCol(length=10)
> carNumber = IntCol()
> active = BoolCol(default=True)
> createdOn = DateCol(default=datetime.now())
> railCategories = RelatedJoin("RailCategory",
> intermediateTable="railCategoryCar",
> joinColumn="railCarId",
> otherColumn="railCategoryId")
>
> Now I want to get a list of rail cars in a category. However, I want
> the rail cars sorted by the name of the rail car.
>
> Here is where I'm at:
>
> # Get the data and add it to the report
> rcats = RailCategory.select(orderBy='name')
> for rcat in rcats:
> printName = rcat.name
> for rc in rcat.railCars:
> print rc.name
Solution N1, in Python:
rcats = RailCategory.select(orderBy='name')
for rcat in sorted(rcats, key=operator.attrgetter('name')):
Solution N2, still in Python:
class RailCategory(SQLObject):
railCars = RelatedJoin("RailCar", intermediateTable="railCategoryCar",
joinColumn="railCategoryId", otherColumn="railCarId",
orderBy="name")
Solution N3, in SQL:
class RailCategory(SQLObject):
railCars = SQLRelatedJoin("RailCar", intermediateTable="railCategoryCar",
joinColumn="railCategoryId", otherColumn="railCarId",
orderBy="name")
Oleg.
--
Oleg Broytmann http://phd.pp.ru/ [EMAIL PROTECTED]
Programmers don't die, they just GOSUB without RETURN.
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