here is a quick code sample:
in a smartgrid:
...
links=dict(table_a=[
lambda row: A('Duplicate',
_class='button',
_onclick='return confirm("Duplicate %s?")'
% row.po_number,
_href=URL('duplicate_po',args=[row.id])),
lambda row: A('Print', _class='button',
_href=URL('print_all',args=[row.id]))]),
def duplicate_po():
...
def print_all():
....
On Fri, Oct 12, 2012 at 1:49 PM, Simon Carr <[email protected]> wrote:
> Hi All,
>
> How do I add an extra link to the SQLFORM.smartgird
>
> As an example I want to add a link or a button "Process File" and have it
> run a controller function that opens and processes an uploaded file.
>
> Thanks
> Simon
>
> --
>
>
>
>
--
