在rails模型中编写大型SQL的更好方法是什么？

Question

采用了大量的后阿雷尔是Rails的提供了糖的代码,我与庞大而复杂的sql语句的查询,我无法做到这一点非常好打交道时遇到问题阿雷尔方法.我喜欢Arel的小东西,但是当它变得混乱时,我更喜欢分开代码.

那么,有什么建议我应该如何处理模型中的大型SQL？就像,我何时应该为此创建SQL视图(因为我看到Rails提供的不是很好,我必须为此创建一个迁移)或在某个文件夹"sqls"中创建任何单独的类,然后从那里调用.

我知道有些人使用<< - SQL表达式

这是我目前的例子:

Question.from(self.questions
                  .select("questions.id")
                  .select("(NOT (questions.last_active_user_id = #{user.id} OR (COALESCE(ss.updated_at > questions.last_active_at, false) OR COALESCE(ds.updated_at > questions.last_active_at, false))))::integer as active")
                  .select("(((NOT((COALESCE(ss.updated_at > questions.created_at, false) OR COALESCE(ds.updated_at > questions.created_at, false))) AND pages.owner_id = questions.user_id) OR (NOT (COALESCE(ss.updated_at > questions.owner_found_important_at, false) OR COALESCE(ds.updated_at > questions.owner_found_important_at, false)) AND owner_found_important_at is not null AND COALESCE(pages.owner_id <> #{user.id}, true))) AND COALESCE(pages.owner_id <> #{user.id}, true) AND (questions.last_active_user_id <> #{user.id}))::integer as owner_active")
                  .select("COALESCE(COUNT(answers.id) = 0, true)::integer as open")
                  .joins("LEFT JOIN seens ss ON questions.slide_id = ss.viewed_id AND ss.viewed_type = 'Slide' AND ss.viewer_id = #{user.id}")
                  .joins("LEFT JOIN seens ds ON questions.document_id = ds.viewed_id AND ds.viewed_type = 'Document' AND ds.viewer_id = #{user.id}")
                  .joins("INNER JOIN documents ON documents.id = questions.document_id")
                  .joins("INNER JOIN lists ON lists.id = documents.list_id")
                  .joins("INNER JOIN pages ON pages.id = lists.page_id")
                  .joins("LEFT OUTER JOIN answers ON answers.question_id = questions.id")
                  .where("questions.reports_count < 2")
                  .group("questions.id, active, owner_active")
                  .as('questions'))
                  .select("SUM(questions.active) as active, SUM(questions.owner_active) as owner_active, SUM(questions.open) as opened, COUNT(questions.id) as total, SUM(CASE WHEN (questions.active > 0 and questions.open > 0) THEN questions.open ELSE 0 END) as opened_active, SUM(CASE WHEN (questions.owner_active > 0 and questions.open > 0) THEN questions.owner_active ELSE 0 END) as opened_active_owner").first

Answer 1

find_by_sql与此处的文档结合使用：

Questions.find_by_sql(<<SQL)
select questions.id
  ...
SQL