Hi Johann,
That is correct but I just wanted to know if I can change the
existing functionality for SQLForm.grid View button. To be precise -
here is an example of what I am trying to do.
Example:
The default builtin "View" button on SQLForm.grid has label "View",
now how can I change its label to say "View Post". Note, here I do not
intend to over-ride the existing controller(function) that
SQLForm.grid's View provides but just change its label. How can we
achieve this? I hope you get what I am doing here :)
Again - If someone could post the complete syntax for SQLForm.grid(All
possible parameters ). It would be really helpful. Thanks
Thanks and Regards, Rahul
(www.flockbird.com - web2py powered)
On Nov 19, 2:13 am, Johann Spies <johann.sp...@gmail.com> wrote:
> On 18 November 2011 09:00, Rahul <rahul.dhak...@gmail.com> wrote:
>
> > Hi Johan, All,
> > Sorry but I am not able to get it. With your code above I get
> > an invalid controller
>
> You are not supposed to use that code exactly as it is. You must adapt it
> for your circumstances:
>
> > > links = [lambda row:
> > A('Edit',_href=URL("controller","edit",args=["update", tablename, a.id]))
>
> Replace "controller" with your controller e.g. 'default'
> Replace 'edit' with your function in the controller referred to.
>
> Regards
> Johann
> --
> Because experiencing your loyal love is better than life itself,
> my lips will praise you. (Psalm 63:3)
