<offtopic>mantra for users seeking help .... post a working model, post a
working model, post a working model</offtopic>
db.define_table('analyses',
Field('info'),
Field('status'),
format='%(info)s')
db.define_table('samples',
Field('analyses_id', 'reference analyses'),
Field('sample_info')
)
you have to manage it at the beginning....something like
def test():
create = True
if request.args(1) == 'samples.analyses_id':
if db.analyses(request.args(2)).status == 'Submitted':
create = False
grid=SQLFORM.smartgrid(db.analyses,linked_tables=['samples'], create=
create)
return dict(grid=grid)
On Monday, January 14, 2013 9:07:58 PM UTC+1, Kostas M wrote:
>
> Thanks for the clarification.
>
> In my case, I have two tables:
>
>
> db.define_table('analyses',Field('info'),Field('status')
> db.define_table('samples',db.analyses,Field('sample_info')
>
>
> and I use:
>
> grid=SQLFORM.smartgrid(db.analyses,linked_tables=['samples'])
>
> I want when the status of an Analysis is 'Submitted', the user to
> not be able to create new, edit, or delete the submitted samples of an
> Analysis. So,I am using the callback functions for the SQLFORM.smartgrid
> options of
>
> editable and deletable, to control these options, eg:
>
> deletable=dict(analyses=is_open, samples=is_open),
>
> Since create is not a callback,
> how, could I control whether the 'Add' button at Samples should be
> available,
> only if the parent Analysis has not yet been submitted?
>
>
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