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():
> 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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>
>
>
>
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