Do you mean like this:
def uploadfile():
filelist = SQLFORM.smartgrid(db.productdatafiles,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]))]),
)
return dict(filelist=filelist)
I tried this, I dont get any errors but I don't see any new buttons or
links on my grid either. Am I doing something wrong?
Simon
On Friday, 12 October 2012 18:49:42 UTC+1, Simon Carr 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
>
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