当包含不起作用时，如何优化查询？答案

【问题标题】：How do I optimize a query when includes won't work?当包含不起作用时，如何优化查询？
【发布时间】：2015-01-09 22:35:58
【问题描述】：

我有这些each 循环：

<% relation_types.each do |relation| %>

            <% if !current_user.memberships.where(relation: relation).empty? %>
                <% current_user.memberships.where(relation: relation).each do |membership| %>
<% end %>

两者都会在每次页面加载时生成此日志：

 User Load (32.1ms)  SELECT  "users".* FROM "users"  WHERE "users"."id" = 1  ORDER BY "users"."id" ASC LIMIT 1
  FamilyTree Load (6.2ms)  SELECT  "family_trees".* FROM "family_trees"  WHERE "family_trees"."user_id" = $1 LIMIT 1  [["user_id", 1]]
   (3.0ms)  SELECT "memberships"."relation" FROM "memberships"  WHERE "memberships"."family_tree_id" = $1  [["family_tree_id", 1]]
   (3.7ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'great_grandmother'  [["user_id", 1]]
   (17.9ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'great_grandfather'  [["user_id", 1]]
   (3.3ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'grandmother'  [["user_id", 1]]
   (1.8ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'grandfather'  [["user_id", 1]]
   (3.5ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'mom'  [["user_id", 1]]
   (14.9ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'dad'  [["user_id", 1]]
   (2.8ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'aunt'  [["user_id", 1]]
   (3.9ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'uncle'  [["user_id", 1]]
  Membership Load (3.4ms)  SELECT "memberships".* FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'uncle'  [["user_id", 1]]
   (2.7ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'sister'  [["user_id", 1]]
   (14.2ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'brother'  [["user_id", 1]]
  Membership Load (3.2ms)  SELECT "memberships".* FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'brother'  [["user_id", 1]]
   (3.7ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'cousin'  [["user_id", 1]]
   (4.3ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'daughter'  [["user_id", 1]]
   (3.8ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'son'  [["user_id", 1]]
   (5.7ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'niece'  [["user_id", 1]]
   (3.0ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'nephew'  [["user_id", 1]]
  Membership Load (3.9ms)  SELECT "memberships".* FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'nephew'  [["user_id", 1]]
   (2.4ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'granddaughter'  [["user_id", 1]]
   (10.7ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'grandson'  [["user_id", 1]]
   (4.7ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'great_granddaughter'  [["user_id", 1]]
   (5.4ms)  SELECT COUNT(*) FROM "memberships"  WHERE "memberships"."user_id" = $1 AND "memberships"."relation" = 'great_grandson'  [["user_id", 1]]

这是Membership 的架构和关联：

# == Schema Information
#
# Table name: memberships
#
#  id             :integer          not null, primary key
#  family_tree_id :integer
#  user_id        :integer
#  created_at     :datetime
#  updated_at     :datetime
#  relation       :string(255)

class Membership < ActiveRecord::Base
  attr_accessible :user_id, :relation

  belongs_to :family_tree
  belongs_to :user

end

如何优化？

【问题讨论】：

标签： ruby-on-rails ruby-on-rails-4 optimization query-optimization

【解决方案1】：

按关系分组怎么样？

在您的控制器中：

@memberships_grouped_by_relations = current_user.memberships.group_by(&:relation)

在你看来：

<% @memberships_grouped_by_relations.each do |relation, memberships| %>
  <% if memberships.any? %>
    <% memberships.each do |membership| %>
    <% end %>
  <% end %>
<% end %>

【讨论】：

【解决方案2】：

为什么不用 1 个查询 预加载所有成员资格：

memberships = current_user.memberships.where(relation: relation_types)

然后稍后使用 ruby 进行所有过滤/选择：

 <% relation_types.each do |relation| %>
   <% memberships_for_relation = memberships.select { |m| m.relation == relation } %>

   <% if !memberships_for_relation.size > 0 %>
     <% memberships_for_relation.each do |membership| %>
  <% end %>

您可能想编写一些帮助程序来使这段代码更好，但这里的主要思想是一次加载所有必需的成员资格。

【讨论】：