Sorry is a little more complex to get the table name involved.
It should be in request.args(1)
table = request.args(1).rpartition('.')[0]
On Thu, Mar 31, 2016 at 12:10 PM, Massimiliano <mbelle...@gmail.com> wrote:
> Maybe you can check the table involved and set links option accordingly.
>
> Something like:
>
> table = request.args(0) or ‘dashboard’
>
> links = []
> if table == ‘label_dashboard’:
> links=[…]
> grid = SQLFORM.smartgrid(db.dashboard,linked_tables=[‘label_dashboard’],
> links=links)
>
>
> Not tested
>
>
> On Wed, Mar 30, 2016 at 6:40 PM, Alessio Varalta <
> alessio.vara...@ethicalsoftware.it> wrote:
>
>> Hi I have a smartgrid with one child and i want a to add links to my
>> child is possibile for example:
>>
>> I have entity dashboard with a child label_dashboard
>>
>> So i create the smartgrid
>>
>> grid = SQLFORM.smartgrid(db.dashboard,linked_tables=['label_dashboard'])
>>
>> Now I want to add links to label_dashboard is possible? When in the grid
>> i go to label_dashboard i want for every row a link button for go to a
>> page..I know how use the link button in a grid but for the smartgrid?
>>
>> --
>> Resources:
>> - http://web2py.com
>> - http://web2py.com/book (Documentation)
>> - http://github.com/web2py/web2py (Source code)
>> - https://code.google.com/p/web2py/issues/list (Report Issues)
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>
>
>
> --
> Massimiliano
>
--
Massimiliano
--
Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
- https://code.google.com/p/web2py/issues/list (Report Issues)
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