oracle sql: 比同国籍人平均年龄大的人

Question

我想列出所有比具有相同国籍的人的平均年龄大的人。涉及三个表：person、passport 和 country。这是我到目前为止所拥有的：

select round(months_between(sysdate, dob) / 12) as age, country.name, person.name, person.surname
from person
join passport on person.pid= passport.pid
join country on passport.cid= country.cid
where round(months_between(sysdate, dob) / 12) > 
(select avg(round(months_between(sysdate, dob) / 12))
from person join passport on person.pid= passport.pid
join country on passport.cid= country.cid);

由于某种原因，我得到了低于人们平均年龄的不良结果。同国籍人平均年龄的select语句为：

SELECT  avg(round(months_between(sysdate, dateofbirth) / 12)) as age, country.name  
from person
join passport on person.personid = passport.personid
join country on passport.countryid = country.countryid
group by country.name;

我在上面的子选择中没有使用 group by 子句。这可以正常工作，但查询仅针对年龄大于平均年龄的人无法正确生成结果。

Answer 1

这是一个很好的分析函数用例：

SELECT * 
  FROM (select round(months_between(sysdate, dob) / 12) as age,               
               country.name country_name,
               person.name person_first_name, 
               person.surname person_last_name,
               avg(round(months_between(sysdate, dob) / 12)) 
                    over (partition by country.name ) avg_for_country
          from person
                join passport on person.pid = passport.pid
                join country on passport.cid = country.cid
       )
 WHERE age > avg_for_country