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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