Rails: 如何根据条件关联模型的字段对模型进行排序？

Question

我有三个模型：

class Unit < ApplicationRecord
  has_many :feedback_units
  has_many :feedbacks, through: :feedback_units
end

class Feedback < ApplicationRecord
  has_many :feedback_units
  has_many :units, through: :feedback_units
end

class FeedbackUnit < ApplicationRecord
  belongs_to :unit
  belongs_to :feedback
end

这就是我想要的：

反馈模型有一个结果字段。如果结果为真或假，则关闭反馈。如果结果为零，则反馈未关闭。我想从未关闭反馈数量最多的单元到最少的单元进行排序。即使单位没有反馈也应该在列表的底部。

例如：

Unit	Number of Closed Feedbacks	Number of Unclosed Feedbacks	Total Number of Feedbacks
Unit_A	14	5	19
Unit_B	35	23	58
Unit_C	112	17	129
Unit_D	0	0	0
Unit_E	36	0	36
Unit_F	0	19	19
Unit_G	0	0	0

应该排序如下：

Unit	Number of Closed Feedbacks	Number of Unclosed Feedbacks	Total Number of Feedbacks
Unit_B	35	23 <-	58
Unit_F	0	19 <-	19
Unit_C	112	17 <-	129
Unit_A	14	5 <-	19
Unit_D	0	0 <-	0
Unit_E	36	0 <-	36
Unit_G	0	0 <-	0

没有未关闭反馈的单元如何排序并不重要。最后三个单元可以排序如下：

Unit	Number of Closed Feedbacks	Number of Unclosed Feedbacks	Total Number of Feedbacks
.	.	.	.
Unit_E	36	0	36
Unit_D	0	0	0
Unit_G	0	0	0

我试过这样做，但没有成功：

Unit.distinct
    .left_joins(:feedbacks)
    .group('feedbacks.result IS NULL')
    .order('feedbacks.result IS NULL ASC, COUNT(feedbacks.result IS NULL)')

这应该在单个查询和 return ActiveRelation 中。

我该怎么做？

Answer 1

Unit.left_joins(:feedbacks).group("units.id").select(
    'units.id,
     sum(case when feedbacks.result is null and feedbacks.id is not null then 1 else 0 end) AS unclosed, 
     sum(case when feedbacks.result is not null then 1 else 0 end) AS closed, 
     sum(case when feedbacks.id is not null then 1 else 0 end) as total').order('unclosed DESC')

这将适用于 PostgreSQL 和 MySql

Rails: 如何根据条件关联模型的字段对模型进行排序？

Rails: How to sort models by an associated model's field with condition?

sql

postgresql

activerecord

ruby-on-rails

ruby-on-rails-6