On Fri, Oct 2, 2009 at 9:39 PM, BioNip <bionic.nip...@gmail.com> wrote:
>
> Consider this code:
>
> > db.define_table('position',
> > Field('name'),
> > Field('sort_order', 'integer'),
> > )
> >
> > db.define_table('person',
> > Field('name'),
> > Field('position', db.position),
>
if you named this field this way, it would be clearer to you (perhaps - I'm
assuming you're a programmer):
Field('position_id', 'reference position')
> )
> >
> > #This sorts by the position.id
> > db(db.person.name=='Bob').select(orderby=db.person.position)
>
Since a foreign key _is_ a reference, and in this case specifically, a
reference to the other table's id - you are sorting this field by "it's
value" - once you see that, you should not be surprised - position_id is,
specifically, an integer which contains the id (key) of another table; you
are sorting integers, so that is what you get.
> >
> > #This fails.
> > db(db.person.name=='Bob').select(orderby=db.person.position.sort_order)
>
Since there is no such thing or structure as
db.person.position_id.sort_order (you see - position_id is an integer, NOT a
structure - it is merely a reference by id!), and you are suggesting an
inner join
>
> I want a list of people, ordered by their position.sort_order, not
> position.id . Does anyone know how this can be accomplished?
>
I'll let you try to work this out - hint: you want to sort the field IN THE
position table (not the person table), and you are suggesting something that
should be reminiscent of an inner join ('inner' being intersection rather
than union, as an outer join would be) of tables - there are examples
showing almost what you want in the manual (instead of the groupby, use
oderby).
Try to figure it out; let us know if you get stuck.
Regards,
- Yarko
> Thanks.
>
>
> -nip
> >
>
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