如何联接表以选择具有多个标签的文章？

Question

I have a standard article, tags, and tag_map tables. I SELECT authors by a single tag as

SELECT author_id, COUNT(*) AS author_articles FROM articles a 
JOIN tag_map b ON a.article_id=b.article_id
JOIN tags c ON b.tag_id=c.tag_id AND c.tag='tag1'
JOIN author_map d ON a.article_id=d.article_id
JOIN authors e ON d.author_id=e.author_id
GROUP BY e.author_id

but how I can SELECT articles having two or multiple tags?

Answer 1

Couple of simple ways.

Join but where the tag is one of the ones required. Count the DISTINCT articles (as otherwise the count will be doubled), but check that the article has 2 tags using a HAVING clause.

SELECT author_id, 
        COUNT(DISTINCT a.article_id) AS author_articles 
FROM articles a 
JOIN tag_map b ON a.article_id=b.article_id
JOIN tags c ON b.tag_id = c.tag_id AND c.tag IN ('tag1', 'tag2')
JOIN author_map d ON a.article_id=d.article_id
JOIN authors e ON d.author_id=e.author_id
GROUP BY e.author_id
HAVING COUNT(DISTINCT c.tag_id) = 2

Or you could just do an extra join.

SELECT author_id, 
        COUNT(*) AS author_articles 
FROM articles a 
JOIN tag_map b ON a.article_id=b.article_id
JOIN tags c1 ON b.tag_id = c1.tag_id AND c1.tag = 'tag1'
JOIN tags c2 ON b.tag_id = c2.tag_id AND c2.tag = 'tag2'
JOIN author_map d ON a.article_id=d.article_id
JOIN authors e ON d.author_id=e.author_id
GROUP BY e.author_id

Answer 2

You need use conditional Count in your HAVING clause. Try this:

SELECT author_id, COUNT(*) AS author_articles FROM articles a 
  JOIN tag_map b ON a.article_id=b.article_id
  JOIN tags c ON b.tag_id=c.tag_id
  JOIN author_map d ON a.article_id=d.article_id
  JOIN authors e ON d.author_id=e.author_id
GROUP BY e.author_id
HAVING COUNT(c.tag_id)>1

Answer 3

use conditional count() in your having

if you can have duplicated tags you use

 HAVING COUNT(CASE WHEN c.tag ='tag1' THEN 1 END) >= 1
    AND COUNT(CASE WHEN c.tag ='tag2' THEN 1 END) >= 1

if tags are unique you can simplify to

 HAVING COUNT(CASE WHEN c.tag IN ('tag1', 'tag2') THEN 1 END) = 2