PSQL请求太慢。如何修复它？

如何使此请求更有效？

可以使用with。将所有查询移动到WITH并稍加修改：将COUNTcid更改为COUNTDISTINCT cid，并完全删除GROUP BY子句。

您可以使用以下方法：

SELECT 
    COUNT(a)/COUNT(s)*100 as aratio, 
    COUNT(b)/COUNT(s)*100 as bratio, 
    COUNT(c)/COUNT(s)*100 as cratio, 
    COUNT(a), 
    COUNT(b), 
    COUNT(c), 
    COUNT(s) 
FROM 
    (SELECT COUNT(cid) as a FROM images WHERE width > height AND category_id = 4 GROUP BY cid) as aq, 
    (SELECT COUNT(cid) as b FROM images WHERE width < height AND category_id = 4 GROUP BY cid) as bq, 
    (SELECT COUNT(cid) as c FROM images WHERE width = height AND category_id = 4 GROUP BY cid) as cq, 
    (SELECT COUNT(cid) as s FROM images WHERE category_id = 4 GROUP BY cid) as sq;

---

此查询不按cid分组，但您可能不需要它。

您确定需要笛卡尔积吗？不，我不喜欢它，但我无法以更好的方式解决此问题。拆分到不同的查询和/或移动到存储过程？不可能，我需要在一个请求中输入。不要告诉我这是不可能的：与一起使用Dit是可能的。将所有查询移动到WITH并稍加修改：将COUNTcid更改为COUNTDISTINCT cid，并完全删除GROUP BY子句。有时GROUP BY的工作速度比DISTINCT快；重写之前值得检查一下。@Andrew Lazarus:在这个特殊情况下不相关。GROUP BY将返回具有唯一值的N行，而COUNTDISTINCT cid将返回具有唯一值count的一行。

SELECT 
    SUM(CASE (width > height) WHEN true THEN 1 ELSE 0 END)::float8*100/COUNT(*) as aratio, 
    SUM(CASE (width < height) WHEN true THEN 1 ELSE 0 END)::float8*100/COUNT(*) as bratio, 
    SUM(CASE (width = height) WHEN true THEN 1 ELSE 0 END)::float8*100/COUNT(*) as cratio, 
    SUM(CASE (width > height) WHEN true THEN 1 ELSE 0 END), 
    SUM(CASE (width < height) WHEN true THEN 1 ELSE 0 END), 
    SUM(CASE (width = height) WHEN true THEN 1 ELSE 0 END), 
    COUNT(*) 
FROM 
    images WHERE category_id = 4;