# Association count sorting **URL:** \ **Category:** Rails\ **Created:** [July 20, 2010, 4:23am UTC](https://www.ruby-forum.com/t/association-count-sorting/191877 "2010-07-20T04:23:29Z")\ **Posts on this page:** 7\ **Page:** 1
**Author:** ![badnaam](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/e36b37/32.png) [@badnaam](https://www.ruby-forum.com/u/badnaam)\ **Post date:** [July 20, 2010, 4:23am UTC](https://www.ruby-forum.com/t/association-count-sorting/191877/1 "2010-07-20T04:23:29Z")
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 ---
**Author:** ![badnaam](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/e36b37/32.png) [@badnaam](https://www.ruby-forum.com/u/badnaam)\ **Post date:** [July 20, 2010, 4:43am UTC](https://www.ruby-forum.com/t/association-count-sorting/191877/2 "2010-07-20T04:43:58Z")
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. ---
**Author:** ![badnaam](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/e36b37/32.png) [@badnaam](https://www.ruby-forum.com/u/badnaam)\ **Post date:** [July 20, 2010, 4:53am UTC](https://www.ruby-forum.com/t/association-count-sorting/191877/3 "2010-07-20T04:53:21Z")
Thanks. > Grandpa has one to many kids and has one pa > No, Grandpa has\_many pas I didnt quite get how you proposed listing the pas (and kid count for each pa) that belong to a certain grandpa On Jul 19, 7:42 pm, Angel Robert M. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) ---
**Author:** ![badnaam](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/e36b37/32.png) [@badnaam](https://www.ruby-forum.com/u/badnaam)\ **Post date:** [July 20, 2010, 5:15am UTC](https://www.ruby-forum.com/t/association-count-sorting/191877/4 "2010-07-20T05:15:35Z")
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 M. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) ---
**Author:** ![badnaam](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/e36b37/32.png) [@badnaam](https://www.ruby-forum.com/u/badnaam)\ **Post date:** [July 20, 2010, 10:24am UTC](https://www.ruby-forum.com/t/association-count-sorting/191877/5 "2010-07-20T10:24:06Z")
On 20 July 2010 03:22, badnaam [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote: > 2 - List of Pas sorted by the number kids each pa has in descending > order > > * * * > > How do I accomplish #2? A crude way (untested), I am sure there are better as I suspect this will call count zillions of times. It should get you going and you can refactor when you want to. grandpa.pas will give you an array of pas, then simply sort this array by the number of kids. sorted = grandpa.pas.sort\_by { |p| p.kids.count } Colin ---
**Author:** ![badnaam](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/e36b37/32.png) [@badnaam](https://www.ruby-forum.com/u/badnaam)\ **Post date:** [July 20, 2010, 5:01am UTC](https://www.ruby-forum.com/t/association-count-sorting/191877/6 "2010-07-20T05:01:32Z")
> 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. ---
**Author:** ![badnaam](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/e36b37/32.png) [@badnaam](https://www.ruby-forum.com/u/badnaam)\ **Post date:** [July 20, 2010, 10:41am UTC](https://www.ruby-forum.com/t/association-count-sorting/191877/7 "2010-07-20T10:41:05Z")
On 20 July 2010 04:14, badnaam [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote: > Here is _AN_ answer if someone else runs into this… Your solution may certainly work… but is a bit “imperative” in coding terms, and does introduce an n+1 problem. Rails offers a feature called “counter\_cache” on associations - which gives you a column in associate keeping track of the amount of “belonging” records. Have a look for it in the manual here: [http://rails.rubyonrails.org/classes/ActiveRecord/Associations/ClassMethods.html](http://rails.rubyonrails.org/classes/ActiveRecord/Associations/ClassMethods.html) If you use it, you can then just do a normal find without the block and order by the counter-cache column, which avoids all the extra SQL counts.