如何改进这个一对多 ActiveRecord 数据模型？

Question

我有一个自制的（不是我自己的）版本控制系统，其数据结构如下：

  create_table "activities", :force => true do |t|
    t.string   "source"
    t.datetime "created_at",       :null => false
    t.datetime "updated_at",       :null => false
    t.integer  "head_revision_id"
  end

  add_index "activities", ["head_revision_id"], :name => "index_activities_on_head_revision_id"
  add_index "activities", ["source"], :name => "index_activities_on_source"

  create_table "activity_revisions", :force => true do |t|
    t.integer  "activity_id"
    t.string   "activity_type"
    t.string   "title"
    t.text     "content"
    t.text     "comment"
    t.integer  "modified_by_id"
    t.datetime "created_at",                      :null => false
    t.datetime "updated_at",                      :null => false
  end

  add_index "activity_revisions", ["activity_id"], :name => "index_activity_revisions_on_activity_id"
  add_index "activity_revisions", ["activity_type"], :name => "index_activity_revisions_on_activity_type"
  add_index "activity_revisions", ["title"], :name => "index_activity_revisions_on_title"

应用程序显示从最新到最旧的活动列表，分页 (will_paginate) 20 到一个页面。这是用于生成列表的查询：

Activity.where(conditions)
        .joins(:head_revision)
        .includes(:head_revision)
        .order('activities.id DESC')

conditions 根据从搜索表单传递的值而有所不同。对于初始列表显示，conditions 为空白。

从表面上看，这个查询很简单，但在执行过程中，对于大型数据集，它的速度非常慢。我们目前有大约 102,000 条活动记录和 512,000 条 activity_revision 记录。在我们的生产服务器上，查询需要将近 2 秒才能提供计数。在开发环境中，这很糟糕。

我觉得数据模型本身就有问题，我希望有人能告诉我一个更好的方法。

编辑：解释在没有条件的基本查询上运行：

mysql> explain SELECT * FROM `activities`  INNER JOIN `activity_revisions` ON `activity_revisions`.`id` = `activities`.`head_revision_id`;
+----+-------------+--------------------+--------+--------------------------------------+---------+---------+--------------------------------------------+--------+-------+
| id | select_type | table              | type   | possible_keys                        | key     | key_len | ref                                        | rows   | Extra |
+----+-------------+--------------------+--------+--------------------------------------+---------+---------+--------------------------------------------+--------+-------+
|  1 | SIMPLE      | activities         | ALL    | index_activities_on_head_revision_id | NULL    | NULL    | NULL                                       | 106590 |       |
|  1 | SIMPLE      | activity_revisions | eq_ref | PRIMARY                              | PRIMARY | 4       | cms_production.activities.head_revision_id |      1 |       |
+----+-------------+--------------------+--------+--------------------------------------+---------+---------+--------------------------------------------+--------+-------+
2 rows in set (0.00 sec)

关于 count(*) 查询：

mysql> explain SELECT count(*) FROM `activities`  INNER JOIN `activity_revisions` ON `activity_revisions`.`id` = `activities`.`head_revision_id`;
+----+-------------+--------------------+--------+--------------------------------------+--------------------------------------+---------+--------------------------------------------+--------+-------------     +
| id | select_type | table              | type   | possible_keys                        | key                                  | key_len | ref                                        | rows   | Extra            |
+----+-------------+--------------------+--------+--------------------------------------+--------------------------------------+---------+--------------------------------------------+--------+-------------     +
|  1 | SIMPLE      | activities         | index  | index_activities_on_head_revision_id | index_activities_on_head_revision_id | 5       | NULL                                       | 106590 | Using index      |
|  1 | SIMPLE      | activity_revisions | eq_ref | PRIMARY                              | PRIMARY                              | 4       | cms_production.activities.head_revision_id |      1 | Using index      |
+----+-------------+--------------------+--------+--------------------------------------+--------------------------------------+---------+--------------------------------------------+--------+-------------     +
2 rows in set (0.00 sec)

Answer 1

我看到您已经索引了几列，这很好。我想说确保查询尽可能高效的最佳方法之一是确保所有处理数据库中查询/检索的conditions，将其对应的列编入索引 .

Answer 2

猜测查询速度慢的原因并不好玩，幸运的是我们不必这样做。

看看http://guides.rubyonrails.org/active_record_querying.html#running-explain，看看你的Activity查询实际上在做什么。

听起来您正在查询 mysql 数据库，因此请查看那些解释结果中的 key。正如 MilesStanfield 所建议的那样，听起来您会发现您没有有效地使用索引。