On Friday, September 13, 2013 1:57:28 PM UTC-7, 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))))])
>
I'm going to guess the "1 given" arg is "row", and you need to use
"filter=row.id" over on the end of your URL().
See below.
> 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]),
>>
>
like this this ^^^ (see above)
>> _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, .......)
>>
>>
>>
Apologies if I've gotten this wrong.
/dps
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