Question

我正在尝试创建一个查询，以从性别平均身高中获取更高的人的名字。

我知道如何获得每个性别的平均身高：

SELECT avg(height), gender from client group by gender

但是我不知道如何将每个人（基于其性别）与此子查询进行比较。

这是我所能做的：

SELECT cname 
FROM (SELECT height, cname, gender
        FROM client AS PplHeight
        HAVING (height > (SELECT avg(height) from client group by gender))) AS AboveAvg

谢谢。

Answer 1

可以加入：

select client.*
from client
join
(
  select gender, avg(height) as avg_height
  from client
  group by gender
) genders on client.gender = genders.gender and client.height > genders.avg_height;

或使用相关子查询：

select *
from client
where height >
(
  select avg(height)
  from client all_clients
  where all_clients.gender = client.gender
);

Answer 2

您可以尝试子查询，如下所示：

select *
from client as c inner join
(
    SELECT avg(height) as avgh, gender
    from client
    group by gender
) as t
on c.gender = t.gender and c.height > t.avgh

Answer 3

您可以在性别和身高更大的对象上内部联接子查询。

SELECT c1.cname
       FROM client c1
            INNER JOIN (SELECT c2.gender,
                               avg(c2.height) height
                               FROM client c2
                               GROUP BY c2.gender) x
                       ON x.gender = c1.gender
                          AND x.height > c1.height;

Answer 4

SELECT PplHeight.cname, PplHeight.gender, PplHeight.height
FROM 
  client AS PplHeight,
  (SELECT gender, avg(height) as avg_height from client group by gender) avg_h
WHERE PplHeight.gender = avg_h.gender
  AND PplHeight.height > avg_h.avg_height

比较单个字段与SQL中其他字段分组的同一字段的平均值

4 个答案: