每位家长的平均条件为

Question

我有父母posts表和孩子votes与posts.id和votes.post_id的关系。我想统计每个职位的平均评分，但只有最新的50票。我知道该怎么做，对于所有投票：

SELECT T1.`title`, (
    SELECT AVG(`vote`) 
    FROM `votes` 
    WHERE `votes`.`post_id` = T1.`id` 
) AS `average` 
FROM `posts` T1 
GROUP BY T1.`id` 

我知道这可能是与子查询：

SELECT T1.`title`, (
    SELECT AVG(`vote`) 
    FROM (
     SELECT `vote` FROM `votes` 
     WHERE `votes`.`post_id` = T1.`id` 
     ORDER BY `votes`.`id` DESC 
     LIMIT 10 
    ) AS T2 
) AS `average` 
FROM `posts` T1 
GROUP BY T1.`id` 

但有错误：Error in query (1054): Unknown column 'T1.id' in 'where clause'。在子查询中不能访问T1别名。有任何想法吗？

http://sqlfiddle.com/#!9/fb9341/2

Answer 1

好吧，你要挑选在每个职位最新的50行的行。这种类型的查询有很多堆栈溢出的答案，主要是在greatest-n-per-group或limit-per-group标签下。例如：How to SELECT the newest four items per category?

一旦您编写了该查询，您可以将其放入像您已经知道如何编写的子查询中，以获得每个帖子的AVG(vote)。

---

回复您的评论：

这就是我的意思是：鉴于您的SQLFiddle数据

SELECT T1.title, AVG(V.vote) AS avg_vote 
FROM posts T1 
JOIN (
    SELECT v1.id, v1.post_id, v1.vote 
    FROM votes v1 
    LEFT OUTER JOIN votes v2 ON v1.post_id = v2.post_id and v1.id < v2.id 
    GROUP BY v1.id 
    HAVING COUNT(*) < 10 
) AS V ON T1.id = V.post_id 
GROUP BY T1.id; 

输出：

+---------+----------+ 
| title | avg_vote | 
+---------+----------+ 
| Title 1 | 5.4000 | 
| Title 2 | 4.2000 | 
+---------+----------+ 

为了帮助在子查询的JOIN ，你应该在列上有votes的索引（post_id,id）。

---

这里的另一个解决方案，而无需唯一列工作：

SELECT T1.title, AVG(V.vote) AS avg_vote 
FROM posts T1 
JOIN (
    SELECT * 
    FROM (
     SELECT v.*, @r := IF(@p = post_id, @r+1, 1) AS rownum, @p := post_id 
     FROM (SELECT @p:=null, @r:=0) AS _init 
     CROSS JOIN votes v 
     ORDER BY v.post_id, v.id DESC 
    ) AS t 
    WHERE t.rownum <= 10 
) AS V ON T1.id = V.post_id 
GROUP BY T1.id; 

输出是一样的事先查询。