单独的子选择以获得多个计数？

Question

我有这个问题：

SELECT  
(SELECT COUNT(media_id) FROM likes WHERE like_state = true AND media_id = ?) AS likes, 
(SELECT COUNT(media_id) FROM likes WHERE like_state = false AND media_id = ?) AS dislikes, 
(SELECT media_views FROM media_thumbnail WHERE media_id = ?) AS views;

它工作正常，但我不确定这是否是最好的方法。有什么建议吗？

我认为这不需要任何解释，但是...它计算有多少喜欢或不喜欢在 like_state（true = 喜欢，false = 不喜欢）的函数中，并返回视图相关视频的计数。

Answer 1

在表likes 上使用两个计数的单个SELECT 肯定比两个单独的SELECT 更快：

SELECT COUNT(    like_state OR NULL) AS likes
     , COUNT(NOT like_state OR NULL) AS dislikes
     ,(SELECT media_views FROM media_thumbnail WHERE media_id = ?) AS views
FROM   likes
WHERE  media_id = ?;

此查询与不相关的子查询之间存在极端情况差异，您的原始查询与两个隐式 CROSS JOIN（@987654332 中的逗号 , @list 几乎等同于 CROSS JOIN) 以及 Gordon 对最后一个子查询的回答为 CROSS JOIN。

• What does [FROM x, y] mean in Postgres?

前两个子查询总是返回一行，所以CROSS JOIN 不能消除结果。 但是第三个查询可能在表media_thumbnail 中找不到给定的media_id 并返回no row。一个CROSS JOIN 会导致整个查询的no row。

SELECT 列表中带有子查询的我的查询将 no row 转换为空值，并且仍然返回结果行。 LEFT JOIN ... ON true 也可以达到同样的效果：

• JOIN (SELECT ... ) ue ON 1=1?
• What is the difference between LATERAL and a subquery in PostgreSQL?

SELECT 中的子查询如果返回多于一行，则会引发异常 - 如果 media_thumbnail.media_id 定义为 UNIQUE 或 PRIMARY KEY，则不会发生这种情况，并且可能不应该发生无论如何。

关于条件计数：

• For absolute performance, is SUM faster or COUNT?

使用FILTER 子句的语法简洁明了：

SELECT COUNT(*) FILTER (WHERE     like_state) AS likes
     , COUNT(*) FILTER (WHERE NOT like_state) AS dislikes
...

有点冗长，但可能更容易阅读。性能基本相同。

• Return counts for multiple ranges in a single SELECT statement

Answer 2

您的查询很好，尽管我更倾向于将其写成：

SELECT SUM( (l.like_state = true)::int ) as likes,
       SUM( (l.like_status = false)::int ) as dislikes,
       MAX(mt.media_views) as media_views
FROM likes l CROSS JOIN
     (SELECT media_views FROM media_thumbnail WHERE media_id = ?
     ) mt
WHERE l.media_id = ?;

这样可以节省一次扫描likes 表的开销。