I just added the syntax you requested: linked_tables=dict(parent=['child'], child=[])
Please help me test it. On Saturday, 22 September 2012 18:55:57 UTC-5, Adi wrote: > > 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... >> >> >> --