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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