If I may rephrase and clarify a question:
Is it possible to add control, through dictionary in (smartgrid)
linked_tables when will links display?
e.g: linked_tables=dict(parent=['child'], child=[''])
wishful outcome: display children links for parent table, but don't display
any related links for child table.
Thanks,
Adnan
On Friday, September 21, 2012 11:58:05 AM UTC-4, Adi wrote:
>
> I'm wondering what to do in this situation? I have self-referencing fields
> in the child table, and due to that smartgrid display links, which
> basically can't do anything.
>
> I'm trying to eliminate them, but not sure what would be a proper way?
>
> Thanks.
>
> Simplified code sample:
> db.define_table('campaign',
> Field('tbl_uuid', length=64, default=lambda:str(uuid.
> uuid4())),
> Field('name','string', label=T('Campaing Name')),
> format='%(name)s',
> )
>
> db.define_table('message',
> Field('tbl_uuid', length=64, default=lambda:str(uuid.
> uuid4())),
> Field('name','string', label=T('Name')),
> Field('campaign_id', 'reference campaign', label=T(
> 'Campaign')),
> Field('action_yes_id', 'reference message)',
> label=T('Action
> Yes'),),
> Field('action_no_id', 'reference message)',
> label=T('Action
> No')),
> migrate=True)
>
>
>
> grid=SQLFORM.smartgrid(Campaign, details=False, links_in_grid=True,
> linked_tables=['message'],
> # linked_tables=dict(campaign=['message'],
> message=['']),
> links=dict(campaign=[lambda row:(_get_messages(row))
> ]),
> )
> Enter code here...
>
>
>
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