I'm now using the .represent property for the table field.
In my model, I added:
db.t_child.f_parent.represent = \
lambda value, row: A(row.f_parent.f_name,
_href=URL('parent_manage', args=['t_parent', 'view', 't_parent',
row.f_parent.id], user_signature=True))
Now, all my child items contain a link to their parent item and all the
grids don't require another button.
On Saturday, January 26, 2013 8:54:31 AM UTC-5, Michael Beller wrote:
>
> I understand (I think) how to create and manage the button links within a
> smart grid but I'd like to make the reference field value a link to it's
> associated record.
>
> For example ...
>
> # parent table
> db.define_table('t_parent', Field('f_name'))
>
> # child table
> db.define_table('t_child', Field('f_name'), Field('f_parent',
> type='reference t_parent'))
>
> Smartgrid for child table would show f_name, f_parent. What is the best
> way to make the f_parent value in the grid a link to view the parent
> record? The link could be to the action/function for the smartgrid
> associated with the t_parent table (with args to view record).
>
> Thanks in advance.
>
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