【发布时间】:2018-04-17 08:31:13
【问题描述】:
我的目标是找到三个有超过 1 条评论且平均评分 >= 4 的医生
目前我正在使用这项服务
class RatingCounterService
def get_three_best_doctors
doctors = find_doctors_with_reviews
sorted_doctors = sort_doctors(doctors)
reversed_hash = reverse_hash_with_sorted_doctors(sorted_doctors)
three_doctors = get_first_three_doctors(reversed_hash)
end
private
def find_doctors_with_reviews
doctors_with_reviews = {}
Doctor.all.each do |doctor|
if doctor.reviews.count > 0 && doctor.average_rating >= 4
doctors_with_reviews[doctor] = doctor.average_rating
end
end
doctors_with_reviews
end
def sort_doctors(doctors)
doctors.sort_by { |doctor, rating| rating }
end
def reverse_hash_with_sorted_doctors(sorted_doctors)
reversed = sorted_doctors.reverse_each.to_h
end
def get_first_three_doctors(reversed_hash)
reversed_hash.first(3).to_h.keys
end
end
这很慢。
我的医生模型:
class Doctor < ApplicationRecord
has_many :reviews, dependent: :destroy
def average_rating
reviews.count == 0 ? 0 : reviews.average(:rating).round(2)
end
end
审查模型:
class Review < ApplicationRecord
belongs_to :doctor
validates :rating, presence: true
end
我可以通过此请求找到所有有超过 1 条评论的医生
doctors_with_reviews = Doctor.joins(:reviews).group('doctors.id').having('count(doctors.id) > 0')
但是,如果“平均评分”是实例方法,我如何找到平均评分 >= 4 的医生并按最高评分排序?
【问题讨论】:
-
为了找到"with 1 or more"评论,所有你需要做的是
Doctor.joins(:reviews)。为了找到 "with more than 1"(这是你写的),你实际上需要做Doctor.joins(:reviews).group('doctors.id').having('count(doctors.id) > 1'). -
But how can I find doctors with an average rating >= 4 and order them by the highest rating if the "average rating" is an instance method?你可以简单地尝试Doctor.includes(:reviews).where("reviews.avarage(:rating).round(2) >= ?", 4).group('doctors.id').having('count(doctors.id) > 0')或在上面创建一个范围并使用范围
标签: ruby-on-rails ruby activerecord