Rails按has_many关联的结果计数排序

Question

无论如何我可以通过子模型()返回的项目数来订购结果(ASC/ )吗？DESCJobs

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

例如:我需要打印最高职位的公司

Answer 1

Rails 5+

引入了对左外连接的支持,Rails 5因此您可以使用外连接而不是使用它counter_cache来执行此操作.这样您仍然可以保留具有0个关系的记录:

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

查询的SQL等价物是这个(通过调用.to_sql它得到):

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

如果您希望经常使用此查询,我建议您使用内置的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

然后像使用它一样

@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

添加到 Tan 的答案中。包括 0 个关联

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

默认情况下，joins使用内连接。我尝试使用left join包含 0 关联