介紹
在Nilenso，哥在搞一個 (開源的哦!)用來設(shè)計和發(fā)起調(diào)查的應(yīng)用。
下面這個是一個調(diào)查的例子：
在內(nèi)部，它是這樣表示滴：
一個調(diào)查包括了許多問題（question）。一系列問題可以歸到（可選）一個分類（category）中。我們實際的數(shù)據(jù)結(jié)構(gòu)會復(fù)雜一點（特別是子問題sub-question部分），但先當(dāng)它就只有question跟category吧。
我們是這樣保存question跟category的。
每個question和category都有一個order_number字段。是個整型，用來指定它自己與其它兄弟的相對關(guān)系。
舉個例子，比如對于上面這個調(diào)查：
Bar的order_number比Baz的小。
這樣一個分類下的問題就能按正確的順序出現(xiàn)：
# In category.rb def sub_questions_in_order questions.order('order_number')end實際上一開始我們就是這樣fetch整個調(diào)查的。每個category會按順序獲取到全部其下的子問題，依此類推遍歷整個實體樹。
這就給出了整棵樹的深度優(yōu)先的順序：
對于有5層以上的內(nèi)嵌、多于100個問題的調(diào)查，這樣搞跑起來奇慢無比。
遞歸查詢
哥也用過那些awesome_nested_set之類的gem，但據(jù)我所知，它們沒一個是支持跨多model來fetch的。
后來哥無意中發(fā)現(xiàn)了一個文檔說PostgreSQL有對遞歸查詢的支持！唔，這個可以有。
那就試下用遞歸查詢搞搞這個問題吧（此時哥對它的了解還很水，有不到位，勿噴）。
要在Postgres做遞歸查詢，得先定義一個初始化查詢，就是非遞歸部分。
本例里，就是最上層的question跟category。最上層的元素不會有父分類，所以它們的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)
(這個查詢和接下來的查詢假定要獲取的是id為2的調(diào)查)
這就獲取到了最上層的元素。
下面要寫遞歸的部分了。根據(jù)下面這個Postgres文檔：
遞歸部分就是要獲取到前面初始化部分拿到的元素的全部子項。
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，遞歸部分只能獲取question。如果一個子項的第一個子分類是個分類呢？Postgres不給引用非遞歸項超過一次。所以在question跟category結(jié)果集上做UNION是不行的。這里得搞個改造一下：
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;
在與非遞歸部分join之前就將category和question結(jié)果集UNION了。
這就產(chǎn)生了所有的調(diào)查元素：
不幸的是，順序好像不對。
在遞歸查詢內(nèi)排序
這問題出在雖然有效的為一級元素獲取到了全部二級元素，但這做的是廣度優(yōu)先的查找，實際上需要的是深度優(yōu)先。
這可怎么搞呢？
Postgres有能在查詢時建array的功能。
那就就建一個存放fetch到的元素的序號的array吧。將這array叫做path好了。一個元素的path就是：
父分類的path（如果有的話）+自己的order_number
如果用path對結(jié)果集排序，就可以將查詢變成深度優(yōu)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;
這很接近成功了。但有兩個 What's your favourite song?
這是由比較ID來查找子項引起的：
WHERE e.category_id = fle.id
fle同時包含question和category。但需要的是只匹配category（因為question不會有子項）。
那就給每個這樣的查詢硬編碼一個類型(type)吧，這樣就不用試著檢查question有沒有子項了：
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;
這看起來就ok了。搞定！
下面就看看這樣搞的性能如何。
用下面這個腳本（在界面上創(chuàng)建了一個調(diào)查之后），哥生成了10個子問題序列，每個都有6層那么深。
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
每個問題序列看起來是這樣滴：
那就來看看遞歸查詢有沒有比一開始的那個快一點吧。
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？不錯不錯。
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