Question

比方说，我需要查找所有带有所有三个标签标记的文章：food，lifestyle和health。在MySQL中最有效的方法是什么？我想出了这个解决方案：

select * from articles
where exists (
    select * from tags
    join article_tag on article_tag.tag_id = tags.id
    where article_tag.article_id = articles.id
    and tags.tag = 'food'
) and exists (
    select * from tags
    join article_tag on article_tag.tag_id = tags.id
    where article_tag.article_id = articles.id
    and tags.tag = 'lifestyle'
) and exists (
    select * from tags
    join article_tag on article_tag.tag_id = tags.id
    where article_tag.article_id = articles.id
    and tags.tag = 'health'
)

它工作正常，但是看起来很重复。解决此问题的最有效的查询是什么？

Answer 1

select a.*
from articles a 
join (
select articles.id
from articles
join article_tag on article_tag.article_id = articles.id
join tags on article_tag.tag_id = tags.id
where tags.tag in ('food','lifestyle','health')
group by articles.id
having SUM(CASE WHEN tags.tag = 'food' THEN 1 ELSE 0 END) >= 1
AND SUM(CASE WHEN tags.tag = 'lifestyle' THEN 1 ELSE 0 END) >= 1
AND SUM(CASE WHEN tags.tag = 'health' THEN 1 ELSE 0 END) >= 1) b on a.id = b.id

Answer 2

您可以改为使用条件聚合：

select a.*
from articles a join
     article_tag at
     on at.article_id = a.id join
     tags t
     on at.tag_id = t.id
where t.tag in ('food', 'lifestyle', 'health')
group by a.id
having count(*) = 3;

这有两个合理的假设：

id是article中的主键（或至少是唯一键）
tag_id没有重复

Answer 3

这是解决方案：

select a.*
from
  articles a
    inner join(
      select at.article_id
      from
        article_tag at
          inner join tags t
          on t.id = at.tag_id
      where t.tag in ('food', 'lifestyle', 'health')
      group by at.article_id
      having count(distinct t.tag) = 3
    ) at
    on at.article_id = a.id

Answer 4

  select * from articles
where articles.id in (
    select article_tag.article_id from tags
    join article_tag on article_tag.tag_id = tags.id
    where article_tag.article_id = articles.id
    and tags.tag in ('food', 'lifestyle', 'health')        
    having count(*) >= 3    
)

检查是否存在多个关系的最有效方法

4 个答案: