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?)'
%
OK Sorted it,
Not sure why your example didn't work, but I took a very similar example
from another web site and it worked fine.
Thanks anyway for putting me on the write track.
Simon
On 12 October 2012 20:35, Simon Carr simonjc...@gmail.com wrote:
Do you mean like this:
def uploadfile():
Sorry, in case anyone else comes across this post looking for the same
thing, here is the code I used to generate my smartgird.
def uploadfile():
filelist = SQLFORM.smartgrid(db.productdatafiles,links = [lambda row:
A(T('Process File'),_href=URL(vital,processfile,args=[row.id]))])
return
Hi Simon,
It didn't work because I put there table_a, referring to a table where
you need your button displayed. In your case, you could put there
productdatafiles (without db in front).
Since smartgrid drills into related tables, some buttons will not be
applicable to those tables, like in my
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