我有这样的架构:
Posts (id, title, created_on)
Categories (id, title, description)
Distributions (id, post_id*, category_id, main)
posts
表格中的每个帖子都有一个或多个类别,只有一个主要类别。
我希望在以下条件下获得每个类别的5个帖子:
所有帖子的年龄必须超过90天
对于每个类别,帖子按DESC中的created_on
排序
这就是我的想法,但无法做到正确:
SELECT p.title, c.title, c.description FROM posts p, distributions d, categories c
WHERE p.id = d.post_id
AND d.category_id = c.id
AND d.main = 1
AND ABS(DATEDIFF(NOW(), p.created_on)) > 90
GROUP BY c.id
ORDER BY p.created_on DESC
LIMIT 5;
我不确定limit 5
,它是返回连接表的前5行还是每个类别组的5行?
任何解释都将不胜感激。谢谢!
答案 0 :(得分:2)
在MySQL中使用会话变量,因为它目前不支持任何分析功能。下面的子查询将为每个category_title
生成序列号,并使用该列在外部查询中进行过滤。
SELECT post_title, category_title, description
FROM
(
SELECT p.title AS post_title,
c.title AS category_title,
c.description
@counter := IF(@current_category = c.title, @counter + 1, 1) AS counter,
@current_category := c.title
FROM posts p, distributions d, categories c
WHERE p.id = d.post_id
AND d.category_id = c.id
AND d.main = 1
AND ABS(DATEDIFF(NOW(), p.created_on)) > 90
ORDER BY p.created_on DESC
) s
WHERE counter <= 5
虽然结构不一样,但是这个DEMO会显示查询的结果。
答案 1 :(得分:-1)
select c.id,c.title,p.id,p.title,description,created_on from Categories as c
join Distributions as d on d.category_id = c.id
join Posts as p on p.id = d.post_id
where ABS(DATEDIFF(NOW(), p.created_on)) > 90
ORDER BY p.created_on DESC
LIMIT 5;