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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