介紹
在Nilenso，哥在搞一個(gè) (開(kāi)源的哦!)用來(lái)設(shè)計(jì)和發(fā)起調(diào)查的應(yīng)用。
下面這個(gè)是一個(gè)調(diào)查的例子：
在內(nèi)部，它是這樣表示滴：
一個(gè)調(diào)查包括了許多問(wèn)題（question）。一系列問(wèn)題可以歸到（可選）一個(gè)分類（category）中。我們實(shí)際的數(shù)據(jù)結(jié)構(gòu)會(huì)復(fù)雜一點(diǎn)（特別是子問(wèn)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(gè)question和category都有一個(gè)order_number字段。是個(gè)整型，用來(lái)指定它自己與其它兄弟的相對(duì)關(guān)系。
舉個(gè)例子，比如對(duì)于上面這個(gè)調(diào)查：
Bar的order_number比Baz的小。
這樣一個(gè)分類下的問(wèn)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(shí)際上一開(kāi)始我們就是這樣fetch整個(gè)調(diào)查的。每個(gè)category會(huì)按順序獲取到全部其下的子問(wèn)題，依此類推遍歷整個(gè)實(shí)體樹(shù)。
這就給出了整棵樹(shù)的深度優(yōu)先的順序：
對(duì)于有5層以上的內(nèi)嵌、多于100個(gè)問(wèn)題的調(diào)查，這樣搞跑起來(lái)奇慢無(wú)比。
遞歸查詢
哥也用過(guò)那些awesome_nested_set之類的gem，但據(jù)我所知，它們沒(méi)一個(gè)是支持跨多model來(lái)fetch的。
后來(lái)哥無(wú)意中發(fā)現(xiàn)了一個(gè)文檔說(shuō)PostgreSQL有對(duì)遞歸查詢的支持！唔，這個(gè)可以有。
那就試下用遞歸查詢搞搞這個(gè)問(wèn)題吧（此時(shí)哥對(duì)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(gè)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(huì)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(gè)查詢和接下來(lái)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(xiě)遞歸的部分了。根據(jù)下面這個(gè)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(xiàng)。
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(gè)子項(xiàng)的第一個(gè)子分類是個(gè)分類呢？Postgres不給引用非遞歸項(xiàng)超過(guò)一次。所以在question跟category結(jié)果集上做UNION是不行的。這里得搞個(gè)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(duì)。
在遞歸查詢內(nèi)排序
這問(wèn)題出在雖然有效的為一級(jí)元素獲取到了全部二級(jí)元素，但這做的是廣度優(yōu)先的查找，實(shí)際上需要的是深度優(yōu)先。
這可怎么搞呢？
Postgres有能在查詢時(shí)建array的功能。
那就就建一個(gè)存放fetch到的元素的序號(hào)的array吧。將這array叫做path好了。一個(gè)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(duì)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(gè) What's your favourite song?
這是由比較ID來(lái)查找子項(xiàng)引起的：
WHERE e.category_id = fle.id
fle同時(shí)包含question和category。但需要的是只匹配category（因?yàn)閝uestion不會(huì)有子項(xiàng)）。
那就給每個(gè)這樣的查詢硬編碼一個(gè)類型(type)吧，這樣就不用試著檢查question有沒(méi)有子項(xiàng)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(lái)就ok了。搞定！
下面就看看這樣搞的性能如何。
用下面這個(gè)腳本（在界面上創(chuàng)建了一個(gè)調(diào)查之后），哥生成了10個(gè)子問(wèn)題序列，每個(gè)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(gè)問(wèn)題序列看起來(lái)是這樣滴：
那就來(lái)看看遞歸查詢有沒(méi)有比一開(kāi)始的那個(gè)快一點(diǎn)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(cuò)不錯(cuò)。
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