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在PostgreSQL中实现递归查询的教程

发布时间:2015-04-21 作者:投稿goldensun 来源:转载
这篇文章主要介绍了在PostgreSQL中实现递归查询的教程,包括在递归查询内排序等方法的介绍,需要的朋友可以参考下

介绍

在Nilenso,哥在搞一个 (开源的哦!)用来设计和发起调查的应用。

下面这个是一个调查的例子:

2015421111930604.png (1436×992)

在内部,它是这样表示滴:

2015421112055942.png (787×751)

一个调查包括了许多问题(question)。一系列问题可以归到(可选)一个分类(category)中。我们实际的数据结构会复杂一点(特别是子问题sub-question部分),但先当它就只有question跟category吧。

我们是这样保存question跟category的。

每个question和category都有一个order_number字段。是个整型,用来指定它自己与其它兄弟的相对关系。

举个例子,比如对于上面这个调查:

2015421112241113.png (482×219)

Bar的order_number比Baz的校

这样一个分类下的问题就能按正确的顺序出现:

# In category.rb
 
def sub_questions_in_order
 questions.order('order_number')
end

实际上一开始我们就是这样fetch整个调查的。每个category会按顺序获取到全部其下的子问题,依此类推遍历整个实体树。

这就给出了整棵树的深度优先的顺序:

2015421112308130.png (999×504)

对于有5层以上的内嵌、多于100个问题的调查,这样搞跑起来奇慢无比。

递归查询

哥也用过那些awesome_nested_set之类的gem,但据我所知,它们没一个是支持跨多model来fetch的。

后来哥无意中发现了一个文档说PostgreSQL有对递归查询的支持!唔,这个可以有。

那就试下用递归查询搞搞这个问题吧(此时哥对它的了解还很水,有不到位,勿喷)。

要在Postgres做递归查询,得先定义一个初始化查询,就是非递归部分。

本例里,就是最上层的question跟category。最上层的元素不会有父分类,所以它们的category_id是空的。

(
 SELECT id, content, order_number, type, category_id FROM questions
 WHERE questions.survey_id = 2 AND questions.category_id IS NULL
)
UNION
(
 SELECT id, content, order_number, type, category_id FROM categories
 WHERE categories.survey_id = 2 AND categories.category_id IS NULL
)

(这个查询和接下来的查询假定要获取的是id为2的调查)

这就获取到了最上层的元素。

2015421112756202.png (615×252)

下面要写递归的部分了。根据下面这个Postgres文档:

2015421112828817.png (883×190)

递归部分就是要获取到前面初始化部分拿到的元素的全部子项。

WITH RECURSIVE first_level_elements AS (
 -- Non-recursive term
 (
  (
   SELECT id, content, order_number, category_id FROM questions
   WHERE questions.survey_id = 2 AND questions.category_id IS NULL
  UNION
   SELECT id, content, order_number, category_id FROM categories
   WHERE categories.survey_id = 2 AND categories.category_id IS NULL
  )
 )
 UNION
 -- Recursive Term
 SELECT q.id, q.content, q.order_number, q.category_id
 FROM first_level_elements fle, questions q
 WHERE q.survey_id = 2 AND q.category_id = fle.id
)
SELECT * from first_level_elements;

等等,递归部分只能获取question。如果一个子项的第一个子分类是个分类呢?Postgres不给引用非递归项超过一次。所以在question跟category结果集上做UNION是不行的。这里得搞个改造一下:

WITH RECURSIVE first_level_elements AS (
 (
  (
   SELECT id, content, order_number, category_id FROM questions
   WHERE questions.survey_id = 2 AND questions.category_id IS NULL
  UNION
   SELECT id, content, order_number, category_id FROM categories
   WHERE categories.survey_id = 2 AND categories.category_id IS NULL
  )
 )
 UNION
 (
   SELECT e.id, e.content, e.order_number, e.category_id
   FROM
   (
    -- Fetch questions AND categories
    SELECT id, content, order_number, category_id FROM questions WHERE survey_id = 2
    UNION
    SELECT id, content, order_number, category_id FROM categories WHERE survey_id = 2
   ) e, first_level_elements fle
   WHERE e.category_id = fle.id
 )
)
SELECT * from first_level_elements;

在与非递归部分join之前就将category和question结果集UNION了。

这就产生了所有的调查元素:

2015421112900874.png (628×342)

不幸的是,顺序好像不对。

在递归查询内排序

这问题出在虽然有效的为一级元素获取到了全部二级元素,但这做的是广度优先的查找,实际上需要的是深度优先。

这可怎么搞呢?

Postgres有能在查询时建array的功能。

那就就建一个存放fetch到的元素的序号的array吧。将这array叫做path好了。一个元素的path就是:

父分类的path(如果有的话)+自己的order_number

如果用path对结果集排序,就可以将查询变成深度优先的啦!

WITH RECURSIVE first_level_elements AS (
 (
  (
   SELECT id, content, category_id, array[id] AS path FROM questions
   WHERE questions.survey_id = 2 AND questions.category_id IS NULL
  UNION
   SELECT id, content, category_id, array[id] AS path FROM categories
   WHERE categories.survey_id = 2 AND categories.category_id IS NULL
  )
 )
 UNION
 (
   SELECT e.id, e.content, e.category_id, (fle.path || e.id)
   FROM
   (
    SELECT id, content, category_id, order_number FROM questions WHERE survey_id = 2
    UNION
    SELECT id, content, category_id, order_number FROM categories WHERE survey_id = 2
   ) e, first_level_elements fle
   WHERE e.category_id = fle.id
 )
)
SELECT * from first_level_elements ORDER BY path;

2015421113222989.png (999×360)

这很接近成功了。但有两个 What's your favourite song?

这是由比较ID来查找子项引起的:

WHERE e.category_id = fle.id

fle同时包含question和category。但需要的是只匹配category(因为question不会有子项)。

那就给每个这样的查询硬编码一个类型(type)吧,这样就不用试着检查question有没有子项了:

WITH RECURSIVE first_level_elements AS (
 (
  (
   SELECT id, content, category_id, 'questions' as type, array[id] AS path FROM questions
   WHERE questions.survey_id = 2 AND questions.category_id IS NULL
  UNION
   SELECT id, content, category_id, 'categories' as type, array[id] AS path FROM categories
   WHERE categories.survey_id = 2 AND categories.category_id IS NULL
  )
 )
 UNION
 (
   SELECT e.id, e.content, e.category_id, e.type, (fle.path || e.id)
   FROM
   (
    SELECT id, content, category_id, 'questions' as type, order_number FROM questions WHERE survey_id = 2
    UNION
    SELECT id, content, category_id, 'categories' as type, order_number FROM categories WHERE survey_id = 2
   ) e, first_level_elements fle
   -- Look for children only if the type is 'categories'
   WHERE e.category_id = fle.id AND fle.type = 'categories'
 )
)
SELECT * from first_level_elements ORDER BY path;

2015421113429449.png (691×342)

这看起来就ok了。搞定!

下面就看看这样搞的性能如何。

用下面这个脚本(在界面上创建了一个调查之后),哥生成了10个子问题序列,每个都有6层那么深。

survey = Survey.find(9)
10.times do
 category = FactoryGirl.create(:category, :survey => survey)
 6.times do
  category = FactoryGirl.create(:category, :category => category, :survey => survey)
 end
 FactoryGirl.create(:single_line_question, :category_id => category.id, :survey_id => survey.id)
end

每个问题序列看起来是这样滴:

2015421113346254.png (448×397)

那就来看看递归查询有没有比一开始的那个快一点吧。

pry(main)> Benchmark.ms { 5.times { Survey.find(9).sub_questions_using_recursive_queries }}
=> 36.839999999999996
 
pry(main)> Benchmark.ms { 5.times { Survey.find(9).sub_questions_in_order } }
=> 1145.1309999999999

快了31倍以上?不错不错。

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