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 > --