祖先：按孩子人数排序

Question

I want to sort my parent object on basis of number of children it has. Ancestory has this method 'child_ids' which return all the children ids. Is there a way I can use that to sort my parent objects?

 Project.order(child_ids.count)

or

 Project.order(Project.child_ids.count)

don't work. I guess that is obvious. Any other way to sort on basis of children can work too. Thanks!

EDIT: look at ancestry column.Project id 2 has two children 6 and 9, both project 1 and 3 has one. I need to sort on basic of no of children, ie 2, 1, 3, rest...

Answer 1

This should definitely be possible, you just need to join the two tables in order to use the COUNT of the Project s children.

Because you didn't include your schema, I made some assumptions about class names and foreign key/primary key names. My solution assumes the child class has a table name of children, primary key of id, and a foreign key of project_id which references the Project class/projects table.

Project.joins("LEFT OUTER JOIN children on projects.id = children.project_id") .group("projects.id") .order("count(children.id) desc")

It's necessary to use a LEFT OUTER JOIN in order to include the Project s with zero children.

(edit: please add the schema for the child class as well as the class definitions of Project and its child class)

EDIT Wed May 20 07:34:58 PDT 2015

If ancestry holds the id of a Project s child Project , a self join would work.
What is the result of Project.find(6).child_ids ?
What is the result of Project.find(7).child_ids ?
I would expect 2 and 3 respectively.

Project.joins("LEFT OUTER JOIN projects p1 on projects.id = p1.ancestry .group("projects.id") .order("count(p1.ancestry) desc")