自定义搜索引擎
您将如何实现自定义搜索引擎？
您对这样的事情有何看法：
SELECT *
FROM jobs
WHERE job_id IN (
SELECT job_id
FROM job_words
WHERE word_id IN (SELECT word_id FROM words w WHERE text = 'carpenter'))
AND job_id IN (
SELECT job_id
FROM job_words
WHERE word_id IN (SELECT word_id FROM words w WHERE text = 'buildings'))
或这样：
SELECT j.*
,s.matches
FROM jobs as j INNER JOIN
(SELECT jw.job_id, count(*) as matches
FROM job_words AS jw
INNER JOIN (SELECT word_id FROM words w WHERE text IN ('carpenter', 'buildings')) AS w ON w.word_id = jw.word_id
GROUP BY jw.job_id) as s ON s.job_id = j.job_id
How would you implement a custom search engine?
What do you think about something like this:
SELECT *
FROM jobs
WHERE job_id IN (
SELECT job_id
FROM job_words
WHERE word_id IN (SELECT word_id FROM words w WHERE text = 'carpenter'))
AND job_id IN (
SELECT job_id
FROM job_words
WHERE word_id IN (SELECT word_id FROM words w WHERE text = 'buildings'))
or this:
SELECT j.*
,s.matches
FROM jobs as j INNER JOIN
(SELECT jw.job_id, count(*) as matches
FROM job_words AS jw
INNER JOIN (SELECT word_id FROM words w WHERE text IN ('carpenter', 'buildings')) AS w ON w.word_id = jw.word_id
GROUP BY jw.job_id) as s ON s.job_id = j.job_id
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评论（2）
您最好提前根据关键字构建表格。你的代码效率很低。
每次运行此代码时，基本上都会运行 O(n(n+n)) 。相反，请提前为所有“木匠”和“建筑物”制作表格，如果搜索查询的表格不存在，则使用您发布的代码。
You'd be better off building the tables from your keywords in advance. Your code is very inefficient.
You're basically running O(n(n+n)) every time you run this code. Instead make the tables for all the 'carpenter' and 'buildings' before hand and if the table for the search query doesn't exist then use the code you posted.