db(db.owner.isdefault=True).select()
should be
db(db.owner.isdefault==True).select()
On Thursday, 7 June 2012 16:48:57 UTC-5, Vincent wrote:
>
> Thanks for the suggestion.
> It returns the following error:
> SyntaxError: keyword can't be an expression
>
> On Thursday, June 7, 2012 4:26:03 PM UTC-5, Derek wrote:
>>
>> Have you tried this:
>>
>> db.define_table('dog',
>> Field('name','string'),Field('owner',db.owner,default=db(db.owner.isdefault=True).select().first()))
>>
>>
>> On Thursday, June 7, 2012 2:20:30 PM UTC-7, Vincent wrote:
>>>
>>> Hi,
>>>
>>> I'm looking for a way to define a default foreign key dynamically.
>>>
>>> For example if I have the following schema:
>>>
>>> db.define_table('owner',Field('name', 'string'),
>>> Field('isdefault','boolean',default=False))
>>> db.people.insert('name'='Bill')
>>> db.people.insert('name'='unknown',isdefault=True)
>>> db.define_table('dog',
>>> Field('name','string'),Field('owner',db.owner,default=????))
>>> db.dog.insert(name='Fido',owner=1)
>>> db.dog.insert(name='mutt') # hopefully points to "unknown" in owner table
>>>
>>> Is there a way I could dynamically check the "owner" table and return
>>> the id of the first row where isdefault==True whenever I define a new "dog"
>>> without an owner?
>>> In the example above the "mutt" row should have owner==2 ("unknown" in
>>> the owner table).
>>>
>>> Trying to put a function as the default argument does not work. I
>>> realize I could set the owner to always be the first row in the table but I
>>> would like something more robust and flexible.
>>>
>>> Thanks
>>> Vincent
>>>
>>>
