try this :
,links = [lambda row:
A(
SPAN(_class='icon magnifier')
,'View Post'
,_class='button',_title='View Post'
,_href=URL(args=["view", db.person, row.id])
)
]
On Monday, November 21, 2011, Rahul wrote:
> 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 <[email protected] <javascript:;>>
> wrote:
> > On 18 November 2011 09:00, Rahul <[email protected] <javascript:;>>
> 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)
>
