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 dict(filelist=filelist)
def processfile():
return dict()
On 12 October 2012 21:15, Simon Carr <simonjc...@gmail.com> wrote:
> 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
>>>
>> --
>>
>>
>>
>>
>
>
--
