Thanks Massimo!
I'm trying to test it, but got this error bellow:
File "/Users/adnan/web2py-trunk/gluon/sqlhtml.py", line 2280, in smartgrid
linked_tables = linked_tables.get(tablename,[])
UnboundLocalError: local variable 'tablename' referenced before assignment
On Sun, Sep 23, 2012 at 5:31 PM, Massimo Di Pierro <
massimo.dipie...@gmail.com> wrote:
> 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(Campaig**n, 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...
>>>
>>>
>>> --
>
>
>
>
--
Thanks,
Adnan
video: http://vimeo.com/24653283
--
