Rails order by results count of has_many association

Question

Is there anyway I can order the results ( ASC / DESC ) by number of items returned from the child model ( Jobs )?

@featured_companies = Company.joins(:jobs).group(Job.arel_table[:company_id]).order(Job.arel_table[:company_id].count).limit(10)

For example: I need to print the Companies with highest jobs on top

Answer 1

Rails 5+

Support for left outer joins was introduced in Rails 5 so you can use an outer join instead of using counter_cache to do this. This way you'll still keep the records that have 0 relationships:

Company
  .left_joins(:jobs)
  .group(:id)
  .order('COUNT(jobs.id) DESC')
  .limit(10)

---

The SQL equivalent of the query is this (got by calling .to_sql on it):

SELECT "companies".* FROM "companies" LEFT OUTER JOIN "jobs" ON "jobs"."company_id" = "companies"."id" GROUP BY "company"."id" ORDER BY COUNT(jobs.id) DESC

Answer 2

If you expect to use this query frequently, I suggest you to use built-in counter_cache

# Job Model
class Job < ActiveRecord::Base
  belongs_to :company, counter_cache: true
  # ...
end

# add a migration
add_column :company, :jobs_count, :integer, default: 0

# Company model
class Company < ActiveRecord::Base
  scope :featured, order('jobs_count DESC')
  # ...
end

and then use it like

@featured_company = Company.featured

Answer 3

就像是：

Company.joins(:jobs).group("jobs.company_id").order("count(jobs.company_id) desc")

Answer 4

@user24359 正确的应该是：

Company.joins(:jobs).group("companies.id").order("count(companies.id) DESC")

Answer 5

Added to Tan's answer. To include 0 association

Company.joins("left join jobs on jobs.company_id = companies.id").group("companies.id").order("count(companies.id) DESC")

by default, joins uses inner join. I tried to use left join to include 0 association

Answer 6

添加到答案中， direct raw SQL已从 rails 6 中删除，因此您需要将 SQL 包装在Arel （如果raw SQL是安全的，则意味着通过安全避免使用用户输入并以此方式避免SQL injection ） .

Arel.sql("count(companies.id) DESC")

Answer 7

Company.where("condition here...")
       .left_joins(:jobs)
       .group(:id)
       .order('COUNT(jobs.id) DESC')
       .limit(10)