oracle sql：比具有相同国籍的人的平均年龄早的人

Question

我想列出所有年龄大于具有相同国籍的人的平均年龄的人。涉及三个表格：人员，护照和国家。这是我到目前为止有：

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); 

出于某种原因，我得到它小于人的平均年龄不想要的结果。对于同一国籍的人的平均年龄的select语句是：

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; 

我用这个没有GROUP BY子句我再选择在上述范围内。这可以正常工作，但只有年龄大于平均年龄的人才能正确地查询结果。

Answer 1

这是一个很好的用例的解析函数：

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