Question

我有以下脚本：

User.includes(:owned_ratings).map{|x| x.owned_ratings.average(:score)}

调用x.owned_ratings.average(:score)会导致n + 1次查询：

   (0.2ms)  SELECT AVG("ratings"."score") FROM "ratings" INNER JOIN "video_chats" ON "ratings"."video_chat_id" = "video_chats"."id" WHERE "video_chats"."user_id" = $1  [["user_id", 4]]
   (0.1ms)  SELECT AVG("ratings"."score") FROM "ratings" INNER JOIN "video_chats" ON "ratings"."video_chat_id" = "video_chats"."id" WHERE "video_chats"."user_id" = $1  [["user_id", 1]]
   (0.1ms)  SELECT AVG("ratings"."score") FROM "ratings" INNER JOIN "video_chats" ON "ratings"."video_chat_id" = "video_chats"."id" WHERE "video_chats"."user_id" = $1  [["user_id", 5]]
   (0.1ms)  SELECT AVG("ratings"."score") FROM "ratings" INNER JOIN "video_chats" ON "ratings"."video_chat_id" = "video_chats"."id" WHERE "video_chats"."user_id" = $1  [["user_id", 7]]
   (0.1ms)  SELECT AVG("ratings"."score") FROM "ratings" INNER JOIN "video_chats" ON "ratings"."video_chat_id" = "video_chats"."id" WHERE "video_chats"."user_id" = $1  [["user_id", 3]]

为什么包含不使用聚合方法？有没有办法解决这个问题？我知道我可以自己实现平均方法并省略问题，但我想确保没有更好的解决方案。

Answer 1

为什么includes无法使用聚合方法？

因为在ruby中重新实现聚合方法是没有意义的，因为数据库服务器可以更快地完成工作。

事实上，如果这是你需要做的，那么准备和执行原始SQL查询可能会更好，这样整个迭代就在数据库中完成，因此避免了往返和N + 1（更不用说了）加载一切）。

Answer 2

使用SQL实现这一点非常简单：

@users = User.select(
  'users.*, AVG(ratings.score) AS users.average_score'
).group_by('ratings.user_id').joins(:ratings)

@users.map(&:average_score)

Answer 3

因为average实际上需要sql查询？尝试类似：

User.includes(:owned_ratings).map{ |x| x.owned_ratings.map(&:score).instance_eval { reduce(:+) / size.to_f } }

从此SO answer得到的数组中的平均值的方法。效率取决于有多少owned_ratings条记录。

哦Sergio Tulentsev是对的，在db方面做得更好

Ruby on Rails平均方法导致n + 1个查询

3 个答案: