Thank you for the reply.
I couldn't make it like you proposed anyway Anthony's answer worked.
Once again thank you.
On Wednesday, June 12, 2013 2:28:01 PM UTC+2, Cliff Kachinske wrote:
>
> Something like
>
> db.sometable.id.represent lambda value, row: db.sometable[value].name
>
> Or
>
> def get_office_name(id)
> return db((db.relationtable.sometable_id==id) &
> (db.relationtable.othertable_id==db.othertable.id)).select(
> db.othertable.name).first().name
>
> db.sometable.id.represent lambda value, row: get_office_name(value)
>
> On Wednesday, June 12, 2013 4:21:05 AM UTC-4, CrC Nommack wrote:
>>
>> Hello, sorry to bother you guys but I can figure it out how I can solve
>> this problem.
>> I have the following tables:
>>
>> ########################################
>> db.define_table('t_unit',
>> Field('f_unit', type='string', label=T('Unit')),
>> auth.signature,
>> format='%(f_unit)s',
>> singular="Unit",
>> migrate=settings.migrate)
>> ########################################
>> db.define_table('t_offices',
>> Field('f_office', type='string', label=T('Office number')),
>> auth.signature,
>> singular="Office",
>> format='%(f_office)s',
>> migrate=settings.migrate)
>> ########################################
>> db.define_table('t_unit_offices',
>> Field('f_unit', 'reference t_unit', label=T('Unit')),
>> Field('f_office', 'reference t_offices', label=T('Office')),
>> auth.signature,
>> singular="Office",
>> format= '%(f_office)s',
>> migrate=settings.migrate)
>>
>> Is a many to many relationship.
>> One unit can have many offices and one office can be shared by more than
>> one unit.
>>
>> I'm displaying it with a smartgrid:
>>
>> assets = SQLFORM.smartgrid(db.t_unit, linked_tables=['t_unit_offices'],
>> csv = False,
>> deletable = True,
>> editable = True,
>> create = True,
>> paginate=100
>> )
>>
>> That works well.
>>
>> But now I need to have the following relationship: one specific office in
>> one unit have many computers (1:n)
>> I created the following table:
>>
>> ########################################
>> db.define_table('t_computers',
>> Field('f_office', 'reference t_unit_offices', label=T('Office')),
>> Field('f_inv_nr', type='string',
>> label=T('Inventory number')),
>> Field('f_stag', type='string', label=T('ST')),
>> auth.signature,
>> format='%(f_inv_nr)s',
>> singular="Computer",
>> plural="Computers",
>> migrate=settings.migrate)
>>
>> and modified the smartgrid definition like that:
>>
>> assets = SQLFORM.smartgrid(db.t_unit, linked_tables=['t_unit_offices',
>> 't_computers'],
>> csv = False,
>> deletable = True,
>> editable = True,
>> create = True,
>> paginate=100
>> )
>>
>> It works well but when I click in the "Hardwares" link it shows the
>> following in the breadcrumbs:
>>
>> Units>A1>Offices>*8*>Hardware
>>
>> How can I manage to get the name of the office instead of the id?
>>
>> Thank you very much for all.
>>
>> Kind regards.
>>
>
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