Question

 doc.id|tag.id|taggings.id|    name

  3 |  3 |  3 | heroku
  3 |  4 |  4 | javascript
  3 |  5 |  5 | html
  4 |  4 |  6 | javascript
  4 |  3 |  7 | heroku
  4 |  5 |  8 | html
  4 |  6 |  9 | swagger

我有这张桌子

我想选择至少具有N个相同标签的元素

所以我们假设用html和heroku标记的文档。

我想要返回doc 3和doc 4 [在此表中它将是表中唯一的两件事。大声笑但仍然！]

Answer 1

这是count和case的一种方式：

select id
from documents
group by id
having count(case when name = 'html' then 1 end) > 0
   and count(case when name = 'heroku' then 1 end) > 0

SQL Fiddle Demo

Answer 2

一般来说，范围将采用以下格式：

class Doc
   has_and_belongs_to_many :tags

   scope :tagged_with(*list), -> {|list| joins(:tags).merge(Tag.named(*list)) }
end

class Tag
   has_and_belongs_to_many :docs

   scope :named(*list), -> {|list| where(name: list) }
end

我不确定where(name: list)是否能正常工作，所以只需用arel替换范围。

scope :named(*list), -> {|list| where(arel_table[:name].in(list)) }

用法：

Doc.tagged_with(:html, :heroku)

Answer 3

select doc_id
from mytable
where name in ('html','heroku')
group by doc_id
having count(distinct name) = 2

Answer 4

据我所知：

SELECT doc_id, count(*) AS ct
FROM   doc_tag
WHERE  tag_id IN (3,4,5)  -- "the same tags"
GROUP  BY 1
HAVING count(*) >= 3;     -- at least N of those

SQL Fiddle.

如何在SQL中选择至少具有N个类似外键的文档

4 个答案: