写一个查询，以获得所有药物的详细信息，其价格高于同类型所有药物的平均价格

Question

正如问题所说，我对此代码有错误：

select *, avg(price) as average 
from drug 
having avg(drug.price) > (select count(price) from drug) 

请帮助我。

Answer 1

由于我不知道您使用的是哪种数据库，因此我在此处写了通用 -

select a.* 
 from drug a
 where a.price > ( select avg(b.price) from drug b where b.type= a.type)

Answer 2

我不确定这是否是性能最佳的解决方案，但这就是我接近它的方式：

WITH avg_by_type AS (
SELECT type, avg(prive) as avg_type
FROM drug
GROUP BY type
)
SELECT *
FROM drug 
INNER JOIN avg_by_type ON avg_by_type.type = drug.type AND avg_by_type.price < drug.price