You can use a lambda:
db.define_table('other_table',
Field('key_id', 'reference key',
requires=IS_IN_DB(db, 'key.id', lambda r: '%s %s (%s)' % (r.
person_id.first_name,
r.
person_id.last_name, r.id))))
Anthony
On Friday, May 23, 2014 5:38:21 AM UTC-4, Louis Amon wrote:
>
> Say I have a table like this:
>
> db.define_table('person', Field('first_name'), Field('last_name'),
> format='%(first_name)s %(last_name)s'
>
> Now if I build a new table:
>
> db.define_table('key', Field('person_id', 'reference person',
> requires=IS_IN_DB(db, db.person, label=db.person._format)))
>
> My new table 'key' will represent its references using 'person' table's
> format.
>
>
> But what if I have another table like this:
>
> db.define_table('other_table', Field('key_id', 'reference key',
> requires=IS_IN_DB(db, db.key, label='%(first_name)s %(last_name)s
> (%(key_id)s)'???)))
>
>
> How do I define the label in 'other_table' so that it shows the 'person'
> that this key is actually referring to ?
>
>
>
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