Question

我有一个简单的多对多查询，我可以搜索：

SELECT article.*, COUNT(article.id) as hits
FROM article
JOIN article_tag ON article.id = article_tag.article_id
JOIN tag ON tag.id = article_tag.tag_id
WHERE (
    tag.title = "Foo"
    OR tag.title = "Bar"
    OR tag.title = "Foo Bar"
    OR article.title LIKE "%Foo%"
    OR article.title LIKE "%Bar%"
    OR article.title LIKE "%Foo Bar%"
)
AND (article.start <= NOW() OR article.start IS NULL)
AND (article.end > NOW() OR article.end IS NULL)
AND (article.type = 'USER')
AND article.enabled = 1
GROUP BY article.id
ORDER BY hits DESC LIMIT 0, 16

搜索效果不错，但是“点击次数”很长。报告不对，所以排序已关闭。

编辑：我想我现在可以更好地描述我的问题了。如果我将第一行更改为：

SELECT article.title, tag.title

并删除GROUP BY和ORDER BY，结果列出匹配的文章标题和所有引用的tag.titles - 所以标签最多（与搜索词匹配或不匹配）的文章产生的点击率最高。

Answer 1

COUNT(article.id)计算article.id不为null的所有记录。这都是记录。所以它与COUNT(*)相同。您正在计算匹配商品标签的数量。

如果您想计算文章数量，可以统计不同的文章ID：

COUNT(DISTINCT article.id)

Answer 2

如果你这样做会怎么样？试一试

SELECT article.*,X.hits
FROM article
JOIN article_tag ON article.id = article_tag.article_id
JOIN tag ON tag.id = article_tag.tag_id
JOIN 
(
select id,COUNT(id) as hits
from article
group by id
) X ON article.id = X.id
WHERE (
    tag.title in('Foo','Bar','Foo Bar')
    OR article.title LIKE "%Foo%"
    OR article.title LIKE "%Bar%"
    OR article.title LIKE "%Foo Bar%"
)
AND (article.start <= NOW() OR article.start IS NULL)
AND (article.end > NOW() OR article.end IS NULL)
AND (article.type = 'USER')
AND article.enabled = 1
ORDER BY X.hits DESC LIMIT 0, 16

MySQL用JOINS对搜索结果进行了排序

2 个答案: