我希望我的子查询返回多个值（= 每个成员），这可能吗？

Question

在我的网站上，人们可以发表文章并喜欢它们。

我想根据每个成员发表的文章数量和获得的点赞数来计算每个成员的级别。

我有一个工作正常的请求，但由于我猜是子查询，一次只针对一个成员。所以现在，我只能在他们的个人资料页面上这样做。

我想请求所有成员为我的主页做一个前 3 名。

您对我如何解决这个问题有任何想法吗？我在开发方面还很新，所以我想我的代码已经不是那么好了。感谢您的回答和建议。

// this code for one person, works fine
SELECT 
      (select (count(l.idArticle)) as nbLikes 
       from likes as l inner join articles as a on l.idArticle = a.idArticle 
       where a.pseudo ="' . $_SESSION['pseudo']. '") 
       + 
       ( select count(pseudo) as nbArticle 
       from articles and pseudo ="' . $_SESSION['pseudo']. '") 
as sumCount

// code i tried for all member but did not work, saying subqueries have more than one row

SELECT 
       (select (count(l.idArticle)/3) as nbLikes 
       from likes as l inner join articles as a on l.idArticle = a.idArticle 
       group by a.pseudo) 
       + 
       ( select count(pseudo) as nbArticle 
       from articles 
       group by pseudo) 
as sumCount

#1242 - 子查询返回多于 1 行

表结构（抱歉，每列的名称为法语）： Table Article in first - and then table for like. They are quite simple I guess.

点赞表非常简单，人们只能点赞或删除点赞，不能不点赞。 表 Like 和 Article 之间的链接是使用 idArticle 建立的。 我也有会员桌，但这里没有用。

Answer 1

这有点脏，但你可以这样实现。鉴于您需要一篇文章才能获得喜欢，您可以LEFT JOIN 对伪名称的点赞结果，按伪名称分组。使用IFNULL()，以防给定帖子没有赞。

SELECT a.pseudo, 
       a.ArticleCount, 
       IFNULL(likes, 0) as LikesCount, 
       IFNULL(likes, 0)/3 + ArticleCount as Score
FROM (
  SELECT IFNULL(COUNT(a.idArticle), 0) as ArticleCount, a.pseudo
  FROM articles a
  GROUP BY a.pseudo
) as a
-- Find all likes that user has gotten
LEFT JOIN (
  SELECT COUNT(l.idLike) as likes, a.pseudo
  FROM likes l
  JOIN articles a
    ON a.idArticle = l.idArticle
  GROUP BY a.pseudo
) as l
on a.pseudo=l.pseudo

• SQLFiddle 显示正在运行的查询：http://sqlfiddle.com/#!9/375b6f/22

Answer 2

你可以通过伪列加入 suqbeury

select t1.pseudo,t1.nbLikes, t2.nbArticle, t1.nbLikes + t2.nbArticle,
from ( 
  select  a.pseudo, count(l.idArticle) as nbLikes 
  from likes as l 
  inner join articles as a on l.idArticle = a.idArticle 
  group by a.pseudo
) t1   
left join  (
  select pseudo, count(pseudo) as nbArticle 
  from articles 
  group by pseudo
) t2 ON t1.pseudo = t2.pseudo