it works correctly Villas, thanks.
how could I turn "Add a comment!" text into a button. Possible syntax
",_class="btn btn-mini"", but where does it go?
thanks,
Alex
On Saturday, September 14, 2013 10:23:00 AM UTC-7, villas wrote:
>
> Maybe this would work using row.idea.id like this...
>
> grid = SQLFORM.grid(query,
> links = [dict(header='Virtual Field',
> body=lambda row: A('Add a comment!', _href=URL(
> 'comment_on_a_suggestion', vars=dict(filter=row.idea.id <http://row.id/>
> )))
> )]
> )
>
>
> You can also now include "virtual fields" in grids, but I haven't needed
> to try that yet.
>
>
>
> On Saturday, 14 September 2013 15:23:49 UTC+1, Alex Glaros wrote:
>>
>> thanks Villas but I receive this error: Row' object has no attribute 'id'
>>
>> perhaps it doesn't know which of the joined "id"s to reference
>>
>> this work-around works because it let's system know which id it is:
>> db.Idea.id.represent = lambda id, r: A('Add a comment!',
>> _href=URL('comment_on_a_suggestion', vars=dict(filter=id)))
>>
>> Alex
>>
>> On Saturday, September 14, 2013 4:56:33 AM UTC-7, villas wrote:
>>>
>>> Try this...
>>>
>>> grid = SQLFORM.grid(query,
>>> links = [dict(header='Virtual Field',
>>> body=lambda row:
>>> A('Add a comment!', _href=URL('comment_on_a_suggestion', vars=dict(filter=
>>> row.id)))
>>> )]
>>> )
>>>
>>>
>>> On Friday, 13 September 2013 21:57:28 UTC+1, Alex Glaros wrote:
>>>>
>>>> I tried this but got "lambda requires 2 args, 1 given" error, plus need
>>>> to be able to pass Idea.id parm to the button. How to do that?
>>>>
>>>> grid = SQLFORM.grid(query,links = [dict(header='Virtual
>>>> Field',body=lambda id, r: A('Add a comment!',
>>>> _href=URL('comment_on_a_suggestion', vars=dict(filter=id))))])
>>>>
>>>> On Friday, September 13, 2013 1:33:35 PM UTC-7, villas wrote:
>>>>>
>>>>> Hope this helps...
>>>>>
>>>>> From the book:
>>>>>
>>>>> links is used to display new columns which can be links to other
>>>>> pages. The links argument must be a list of
>>>>> dict(header='name',body=lambda
>>>>> row: A(...)) where header is the header of the new column and body is
>>>>> a function that takes a row and returns a value. In the example, the
>>>>> value
>>>>> is a A(...) helper.
>>>>>
>>>>> Example:
>>>>>
>>>>> linkbtns = [
>>>>> lambda row: SPAN('Mag',_class="label label-success")
>>>>> \
>>>>> if row.status =='Y' else '',
>>>>> lambda row: SPAN('Web',_class="label label-success")
>>>>> \
>>>>> if row.is_active else '',
>>>>> lambda row: A( I('',_class="icon-eye-open")+' View',
>>>>> _href=URL("view",args=[row.id]),
>>>>> _class="btn btn-small"
>>>>> ),
>>>>> lambda row: A( I('',_class="icon-edit")+' Edit',
>>>>> _href=URL("edit",args=[row.id]),
>>>>> _class="btn btn-small"
>>>>> ),
>>>>> lambda row: A( I('',_class="icon-road")+' Map',
>>>>> _href=URL('default','geocoder',
>>>>> args=['caclient',row.id],vars={'title':row.business}),
>>>>> _class="btn btn-small"
>>>>> ),
>>>>> ]
>>>>> grid = SQLFORM.grid(....... links=linkbtns, .......)
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> On Friday, 13 September 2013 21:00:56 UTC+1, Alex Glaros wrote:
>>>>>>
>>>>>> How do I add a button "Add a comment!" in every row of a grid? It's a
>>>>>> kind of virtual column.
>>>>>>
>>>>>> The below works, it takes user exactly to the correct row in the next
>>>>>> table, but I use up one of the fields (Idea.id). How do I just create
>>>>>> text/virtual column that doesn't sacrifice existing field?
>>>>>>
>>>>>>
>>>>>> def view_all_suggestions_and_comments():
>>>>>> db.Idea.id.represent = lambda id, r: A('Add a comment!',
>>>>>> _href=URL('comment_on_a_suggestion', vars=dict(filter=id)))
>>>>>> query = (db.IdeaComment.ideaID==db.Idea.id) &
>>>>>> (db.IdeaComment.partyID==db.Party.id)
>>>>>> grid = SQLFORM.grid(query)
>>>>>> return dict(grid = grid)
>>>>>>
>>>>>>
>>>>>> I read the book but need exact syntax for this situation.
>>>>>>
>>>>>> Thanks,
>>>>>>
>>>>>> Alex Glaros
>>>>>>
>>>>>
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