Here is the answer if someone else runs into this..
GrandPa.all do |gpa|
p gpa.name
gpa.pas.all(:joins => :kids, :select => "pas.*, count(pas.id) AS
kid_count"
:group => :id, :order => "kid_count DESC").each do |pa|
p "#{pa.name} : #{pa.kid_count}"
end
end
On Jul 19, 7:58 pm, Angel Robert Marquez <angel.marq...@gmail.com>
wrote:
> > Thanks.
> > > Grandpa has one to many kids and has one pa
> > No, Grandpa has_many pas
>
> How is this possible? That's like saying I have two left arms.
>
>
>
> > I didnt quite get how you proposed listing the pas (and kid count for
> > each pa) that belong to a certain grandpa
>
> Because the sort is to easy to explain. What you're sorting and why is the
> problem. So, now that I have two left arms how do I sort them by thumb
> length on my right arm. Well, I guess get a mirror.
>
>
>
> > On Jul 19, 7:42 pm, Angel Robert Marquez <angel.marq...@gmail.com>
> > wrote:
> > > Kid has one pa and has one grandpa through pa
> > > Pa has one to many kids and has one grandpa
> > > Grandpa has one to many kids and has one pa
>
> > > <%= :name[gp]
>
> > > ?pa = kid if from gp?
>
> > > I think you should just have one model with a foreign key to parent and
> > > quantity of child including nil.
>
> > > On Mon, Jul 19, 2010 at 7:22 PM, badnaam <asitkmis...@gmail.com> wrote:
> > > > I have 3 models GrandPa, Pa, Kid
>
> > > > GrandPa => has_many :pas
> > > > Pa => has_many kids
>
> > > > When I list GrandPa I would like to present the following
>
> > > > *********************************************
> > > > 1 - GrandPa_Name
>
> > > > 2 - List of Pas sorted by the number kids each pa has in descending
> > > > order
> > > > **********************************************
>
> > > > How do I accomplish #2?
>
> > > > Thanks
>
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