我应该如何设计表格以将标签存储在数据库中？

Question

"question_id": 58640
"tags": ["polls", "fun", "quotes"]
"title": "Great programming quotes"
"question_id": 184618
"tags": ["polls", "fun", "comment"]
"title": "What is the best comment in source code you have ever encountered?"
"question_id": 3734102
"tags": ["c++", "linux", "exit-code"]
"title": "Why cant' I return bigger values from main function ?"
"question_id": 2349378
"tags": ["communication", "terminology", "vocabulary"]
"title": "New programming jargon you coined?"
"question_id": 3723817
"tags": ["open-source", "project-management", "failure", "fail"]
"title": "How to make an open source project fail"
"question_id": 3699150
"tags": ["testing", "interview-questions", "job-interview"]
"title": "Interview question please help"

这只是我使用SO API得到的一些问题的文本摘录。

为了使这个可查询，我想使用SQLite 来存储数据。

我应该如何存储 tags 列？

由于这里对 SO 的限制是五个标签，我可以使用五列 tag1、tag2 ...，但我认为可以做一些更优雅的事情。可以扩展到任意数量的标签，并且还可以处理基本查询，例如

select title from table where tag has "c++" and "boost" but not "c"

Answer 1

这是多对多的关系：问题有多个标签，标签可以出现在多个问题中。这意味着您必须创建三个表，一个用于问题，一个用于标签，一个用于这些表之间的链接。结果查询如下所示：

SELECT title FROM question
       INNER JOIN question_tag_link USING (question_id)
       INNER JOIN tag USING (tag_id)
            WHERE tag_name IN('c++', 'boost')
              AND NOT EXISTS(
           SELECT * FROM tag t1
            WHERE t1.tag_name = 'c'
              AND t1.question_id = question.question_id);

没那么简单，但我认为如果你不想被限制，这是要付出的代价。如果少于 64 个不同的标签，您可以使用 SET 字段类型，但您会失去很多灵活性（很难添加新标签）。

Answer 2

select distinct a.QuestionTitle
from
(
select q.QuestionID, QuestionTitle, TagName 
from QuestionTags as x
join Question     as q on q.QuestionID = x.QuestionID 
join Tag          as t on t.TagID      = x.TagID 
where TagName in ('c++', 'boost')
) as a
left join
(
select q.QuestionID, QuestionTitle, TagName 
from QuestionTags as x
join Question     as q on q.QuestionID = x.QuestionID 
join Tag          as t on t.TagID      = x.TagID 
where TagName = 'c'
) as b on b.QuestionID = a.QuestionID
where b.QuestionTitle is null
order by a.QuestionTitle ;