Another way:
tc = t.with_alias('chairs')
q = (tc.name == 'chair')
q = q & (t.name == 'table')
q = q & (t.owner_id == tc.owner_id)
q = q & (p.id == tc.owner_id)
print db(q).select(*fields, distinct=True, orderby=p.first_name)
On Sat, May 9, 2015 at 9:24 PM, Spokes <[email protected]> wrote:
> Thanks, Massimiliano. I was also considering an approach involving
> creating a list of IDs from the 'thing' table, but I'm not sure that this
> would be feasible in this case. They may be tens, possibly hundreds of IDs
> corresponding to the given criteria, and the grid based on these queries is
> part of the site's front page (i.e. it will be generated very often), so
> I'm not sure that this approach is necessarily efficient enough. I can't
> help thinking that there's a more straightforward and efficient way to
> accomplish this task.
>
>
> On Saturday, May 9, 2015 at 8:14:51 AM UTC-5, Massimiliano wrote:
>>
>> Maybe something like that?
>>
>> db.define_table('person', Field('first_name'), Field('last_name'),
>> format='%(first_name)s')
>> db.define_table('thing', Field('name'),
>> Field('owner_id', 'reference person'),
>> format='%(name)s')
>>
>> t = db.thing
>> p = db.person
>>
>> fields = [p.first_name, p.last_name]
>>
>> q = ((t.owner_id == p.id) & (t.name == 'chair'))
>> ids_people_with_chair = [a.id for a in db(q).select(p.id)]
>>
>> query = (t.owner_id == p.id)
>> query = query & (t.owner_id.belongs(ids_people_with_chair))
>> query = query & (t.name == 'table')
>>
>> print db(query).select(*fields, distinct=True, orderby=p.first_name)
>>
>>
>>
>> On Fri, May 8, 2015 at 10:49 PM, Spokes <[email protected]> wrote:
>>
>>> Hi, Dave. Yes, that seems to be an accurate summary of what I was
>>> attempting to do.
>>>
>>>
>>>
>>> On Friday, May 8, 2015 at 3:25:24 PM UTC-5, Dave S wrote:
>>>>
>>>>
>>>>
>>>> On Friday, May 8, 2015 at 11:55:13 AM UTC-7, Spokes wrote:
>>>>>
>>>>> I'd like to create a grid based on criteria relating to a primary
>>>>> table, and a table that references that primary table. Slightly modifying
>>>>> the example from the web2py docs, let's say the primary table is
>>>>> 't_person', and it's referenced by the table, 't_thing'. I'd like the grid
>>>>> to list 't_person' entries that are referenced by a 't_thing' entry with a
>>>>> value 'table' for the 'name' field, and by a 't_thing' entry with value
>>>>> 'chair' for the name field (that is, entries that meet both criteria, not
>>>>> one or the other).
>>>>>
>>>>>
>>>> Let's see if I understand this. You want to use the set A of t_things
>>>> that have value 'table' and the set B of t_things that have value 'chair'
>>>> and find those members of t_person that are referenced by both sets? Sort
>>>> of "for a in A where a.owner_id == b.owner_id for b in B, select p from
>>>> t_person where p.id == a.owner_id ?
>>>>
>>>> /dps
>>>>
>>>> --
>>> Resources:
>>> - http://web2py.com
>>> - http://web2py.com/book (Documentation)
>>> - http://github.com/web2py/web2py (Source code)
>>> - https://code.google.com/p/web2py/issues/list (Report Issues)
>>> ---
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>>> an email to [email protected].
>>> For more options, visit https://groups.google.com/d/optout.
>>>
>>
>>
>>
>> --
>> Massimiliano
>>
> --
> Resources:
> - http://web2py.com
> - http://web2py.com/book (Documentation)
> - http://github.com/web2py/web2py (Source code)
> - https://code.google.com/p/web2py/issues/list (Report Issues)
> ---
> You received this message because you are subscribed to the Google Groups
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>
--
Massimiliano
--
Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
- https://code.google.com/p/web2py/issues/list (Report Issues)
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